Functions and Their Graphs — Free Pre-Calculus Review Games.

The denominator is zero at x = 2, so exclude that value.

\(x^2\) is always non-negative, so the range is \(y \geq 0\).

f(-2) = -2 + 3 = 1.

A graph represents a function if every vertical line hits it at most once.

Adding to the output shifts the graph vertically upward.

Replacing x with (x-2) shifts the graph right by 2.

\(g(3) = 9\), \(f(9) = 2(9)-1 = 17\).

\(3 - x \geq 0\) means \(x \leq 3\).

Negating the output reflects the graph over the x-axis.

\(x = -3 < 0\), so use \(x^2\): \((-3)^2 = 9\).

\(g(f(x)) = (x+1)^2 - 1 = x^2+2x+1-1 = x^2+2x\).

y = (x-3)/2, swap: x = (y-3)/2, 2x = y-3, y = 2x+3.

Even functions satisfy f(-x) = f(x), symmetric about the y-axis.

\(f(x) = x^3\) satisfies \(f(-x) = -f(x)\), making it odd.

Multiplying the output by 2 stretches the graph vertically.

The expression under a square root must be non-negative, so \(x + 5 \geq 0\) gives \(x \geq -5\). The choice "\(x \geq 5\)" is wrong because it incorrectly solves the inequality by ignoring the sign of the constant. Always set the radicand greater than or equal to zero when finding the domain of a square root function.

Since the parabola opens downward with vertex at \((0,4)\), the maximum \(y\)-value is 4 and all outputs are less than or equal to it. The choice "\(y \geq 4\)" is wrong because it describes the range of an upward-opening parabola, not this downward one. Remember that the sign of the leading coefficient determines whether the vertex is a maximum or minimum.

Substituting \(x = 4\) gives \(f(4) = 3(4) - 5 = 12 - 5 = 7\). The choice "\(12\)" is wrong because it stops after multiplying and forgets to subtract 5. Always complete every arithmetic step when evaluating a function at a given input.

Adding a constant inside the function argument shifts the graph horizontally in the opposite direction of the sign, so \(f(x+4)\) moves left 4 units. The choice "Shift right 4 units" is wrong because that would correspond to \(f(x - 4)\), not \(f(x + 4)\). Remember that horizontal shifts behave opposite to their apparent sign inside the parentheses.

The denominator \(x^2 - 9\) equals zero when \(x = 3\) or \(x = -3\), so these values must be excluded from the domain. The choice "All real numbers except \(9\)" is wrong because it confuses the constant term with the actual roots of the factored denominator. When finding domains of rational functions, always factor the denominator and exclude its zeros.

The principal square root function only outputs non-negative values, so its range is \(y \geq 0\). The choice "\(y > 0\)" is wrong because it excludes zero, but \(f(0) = 0\) is a valid output. Recall that the square root function always produces values greater than or equal to zero, including zero itself.

Subtracting a constant outside the function shifts the entire graph vertically downward by that amount. The choice "Shift up 6 units" is wrong because that would result from \(f(x) + 6\), not \(f(x) - 6\). Vertical shifts move directly in the direction of the sign added outside the function.

First evaluate \(g(0) = 0 + 5 = 5\), then substitute into \(f\) to get \(f(5) = 5^2 = 25\). The choice "\(5\)" is wrong because it stops after finding \(g(0)\) and never applies \(f\) to that result. Composition always requires evaluating the inner function first, then applying the outer function to that output.

Because the square root is in the denominator, the radicand must be strictly positive (not just non-negative) so the expression is defined and never divides by zero, giving \(x > 4\). The choice "\(x \geq 4\)" is wrong because at \(x = 4\) the denominator becomes zero, making the function undefined. When a radical appears in a denominator, exclude the value that makes the radicand exactly zero.

Since \(x = 0\) satisfies the condition \(x \geq 0\), we use the second rule: \(f(0) = 0 + 1 = 1\). The choice "\(0\)" is wrong because it mistakenly applies the first rule \(2x\), which only applies when \(x < 0\). Always check which condition the input satisfies before selecting the correct piece of a piecewise function.

A constant function accepts any real number input, so its domain is all real numbers, but it always outputs the single value 7, giving a range of \(\{7\}\). The choice "Domain: \(\{7\}\); Range: all real numbers" is wrong because it reverses the roles of domain and range for a constant function. Constant functions always have an unrestricted domain paired with a single-element range.

Replacing \(x\) with \(-x\) flips every input's sign, which reflects the graph horizontally across the \(y\)-axis. The choice "Reflection over the \(x\)-axis" is wrong because that transformation is represented by \(-f(x)\), which negates the output rather than the input. Distinguish carefully between negating the input (reflect over \(y\)-axis) and negating the output (reflect over \(x\)-axis).

We need \(9 - x^2 \geq 0\), which factors to \((3-x)(3+x) \geq 0\), giving the closed interval \([-3, 3]\) where the expression stays non-negative. The choice "\((-3, 3)\)" is wrong because it excludes the endpoints, even though \(9 - x^2 = 0\) at \(x = \pm 3\), which is allowed since the square root of zero is defined. When solving quadratic inequalities under a square root, test the sign of the factored expression across all intervals including endpoints.

Replacing \(x\) with \(x - 2\) inside the absolute value moves the vertex of the graph from \((0,0)\) to \((2,0)\), a shift to the right by 2 units. The choice "Shift left 2 units" is wrong because that would correspond to \(|x + 2|\), not \(|x - 2|\). Remember horizontal shifts move opposite to the apparent sign written inside the function's argument.

First compute \(g(-1) = -1 + 1 = 0\), then substitute into \(f\): \(f(0) = 2(0)^2 - 3 = -3\). The choice "\(5\)" is wrong because it likely results from evaluating \(g(f(-1))\) instead, mixing up the composition order. The order of composition matters greatly, so always evaluate the innermost function first as indicated by the notation.

The absolute value expression \(|x+3|\) has a minimum value of 0, so the vertex of the shifted graph occurs at \(y = -4\), and all outputs are greater than or equal to \(-4\). The choice "\(y \geq 4\)" is wrong because it incorrectly treats the vertical shift as adding 4 rather than subtracting it. When finding the range of a transformed absolute value function, locate the vertex by identifying the minimum of the base function shifted vertically.

Since \(4 > 1\), we use the rule \(3x - 2\), giving \(f(4) = 3(4) - 2 = 10\). The choice "\(15\)" is wrong because it incorrectly applies \(x^2 - 1\), which is 15 when \(x=4\), but that piece only applies for \(x \leq 1\). Always confirm which interval the input belongs to before applying the corresponding piece of a piecewise-defined function.

The \(+2\) inside the parentheses shifts the graph left 2 units, and the \(-5\) outside shifts it down 5 units, giving the combined transformation. The choice "Shift right 2 units and up 5 units" is wrong because it reverses the direction of both the horizontal and vertical shifts. Remember that inside shifts act opposite to their sign while outside shifts act in the direction of their sign.

The radicand requires \(x \geq -2\), and the denominator requires \(x \neq 3\), so combining these gives $[-2, 3) \cup (3, \infty)$. The choice "$[-2, \infty)$" is wrong because it fails to exclude \(x = 3\), where the function is undefined due to division by zero. When a function has both a radical and a rational expression, address each restriction separately, then combine them.

The composition \((f \circ g)(x) = \sqrt{x - 4}\) requires the radicand to be non-negative, so \(x - 4 \geq 0\) gives \(x \geq 4\). The choice "\(x \geq 0\)" is wrong because it applies the restriction on \(f\)'s original input directly to \(x\), ignoring that \(g(x)\) has shifted the expression under the root. When finding the domain of a composed function, substitute the inner function first and then apply the restriction to the resulting expression.

For \(x<0\), \(x^2\) produces values approaching but not including 0 up to positive infinity, and for \(x \geq 0\), \(-x+2\) produces values from 2 down to negative infinity, so together the outputs cover all real numbers. The choice "\(y \leq 2\)" is wrong because it ignores the first piece, which contributes arbitrarily large positive values as \(x\) becomes very negative. When finding the range of a piecewise function, analyze each piece separately over its restricted domain, then union the results.

Multiplying the entire function by a constant between 0 and 1 compresses the graph vertically toward the \(x\)-axis, so \(\frac{1}{2}f(x)\) is a vertical compression. The choice "Vertical stretch by a factor of \(\frac{1}{2}\)" is wrong because stretches occur when the multiplier's absolute value is greater than 1, not less than 1. A coefficient outside the function between 0 and 1 always compresses vertically, while a coefficient greater than 1 stretches vertically.

First compute \(g(2) = 2^2 + 1 = 5\), then substitute into \(f\): \(f(5) = \sqrt{5-1} = \sqrt{4} = 2\). The choice "\(5\)" is wrong because it only computes the inner function and stops before applying \(f\) to that result. Always finish both steps of a composition problem, evaluating the inner function completely before substituting into the outer function.

The two pieces together cover \(0 \leq x < 4\) and \(x \geq 4\), and combining these intervals gives all \(x \geq 0\) with no gaps. The choice "\(0 \leq x < 4\)" is wrong because it only accounts for the first piece and ignores the second piece that extends the domain to infinity. When determining the domain of a piecewise function, take the union of the domains specified for each individual piece.

A function is even if \(f(-x) = f(x)\), and since \(f(-x) = (-x)^4 - 3(-x)^2 + 1 = x^4 - 3x^2 + 1 = f(x)\), this function satisfies that condition because it contains only even powers of \(x\). The choice "\(f(x) = x^3 - x\)" is wrong because it contains only odd powers of \(x\), making it an odd function rather than an even one. To quickly identify even functions, check whether every exponent of \(x\) in the expression is even, including the constant term treated as \(x^0\).

Multiplying \(x\) by a constant greater than 1 inside the function compresses the graph horizontally toward the \(y\)-axis, so \(f(2x)\) represents a horizontal compression by a factor of \(\frac{1}{2}\). The choice "Horizontal stretch by a factor of 2" is wrong because stretches happen when the coefficient inside the parentheses is between 0 and 1, not greater than 1. Remember that transformations applied to the input \(x\) behave inversely compared to their numerical value, unlike transformations applied outside the function.

Since \(\sqrt{x} \geq 0\), multiplying by \(-2\) makes the term \(-2\sqrt{x} \leq 0\), and adding 3 shifts the maximum possible output to 3, so the range is \(y \leq 3\). The choice "\(y \geq 3\)" is wrong because it fails to account for the negative coefficient, which flips the inequality direction from what a standard square root function would have. When a function is scaled by a negative number, remember that it reverses the direction of the inequality describing its range.

Since \((f \circ g)(x) = 2g(x) + 1 = 4x^2 + 1\), solving for \(g(x)\) gives \(2g(x) = 4x^2\), so \(g(x) = 2x^2\). The choice "\(x^2\)" is wrong because substituting it into \(f\) gives \(2x^2 + 1\), which does not match the required composed result of \(4x^2 + 1\). When working backward to find an inner function from a known composition, set up and solve an equation using the outer function's definition.

Logarithmic functions require their argument to be strictly positive, so \(x - 5 > 0\) gives \(x > 5\). The choice "\(x \geq 5\)" is wrong because at \(x = 5\) the argument becomes zero, and the logarithm of zero is undefined. Always remember that logarithms are only defined for strictly positive arguments, never zero or negative numbers.

The equation \(f(x+2) = f(x)\) means the function's output repeats identically every time \(x\) increases by 2, which is the defining characteristic of a periodic function with period 2. The choice "It is symmetric about the \(y\)-axis" is wrong because that describes even functions, which satisfy \(f(-x) = f(x)\), a completely different condition than periodicity. Recognize that periodicity conditions like \(f(x+c) = f(x)\) indicate the graph's pattern repeats at regular intervals along the \(x\)-axis.

Setting the denominator \(x + 3 = 0\) gives \(x = -3\), and since this factor does not cancel with the numerator, the graph has a vertical asymptote at that value. The choice "\(x = -3\); the graph has a horizontal asymptote there" is wrong because vertical asymptotes arise from zeros of the denominator, whereas horizontal asymptotes come from the degrees and leading coefficients of numerator and denominator. Always find domain restrictions of rational functions by setting the denominator equal to zero and solving.

Swapping \(x\) and \(y\) in \(y = x^2 - 4\) gives \(x = y^2 - 4\), and solving for \(y\) (taking the positive root since the original domain is \(x \geq 0\)) yields \(y = \sqrt{x+4}\); the domain of the inverse equals the range of the original, which is \(x \geq -4\). The choice "\(f^{-1}(x) = \sqrt{x-4}\), domain \(x \geq 4\)" is wrong because it incorrectly reverses the sign inside the radical during the algebraic solving step. When finding an inverse function's domain, remember it always equals the range of the original function.

Evaluating the limit from the left using \(x^2\) as \(x \to 2^-\) gives 4, and evaluating the actual function value using \(3x-2\) at \(x=2\) gives \(3(2)-2=4\), so both sides match and continuity holds. The choice "Yes, because piecewise functions are always continuous" is wrong because piecewise functions are not automatically continuous; continuity must be verified by checking that the pieces meet at the boundary point. To test continuity of a piecewise function at a boundary, always compare the left-hand limit, right-hand limit, and the function's defined value at that point.

Computing \((f \circ g)(x) = 3\left(\frac{x+2}{3}\right) - 2 = x+2-2 = x\) confirms that \(g\) undoes \(f\), satisfying the defining property of inverse functions. The choice "\(f\) and \(g\) are unrelated functions" is wrong because the composition explicitly returns the identity function \(x\), which is the hallmark test for verifying an inverse relationship. To confirm two functions are inverses, always check that composing them in either order returns the input unchanged.

Substituting \(f(x)\) into \(g\) gives \(g(f(x)) = (\sqrt{x-1})^2 + 1 = (x-1) + 1 = x\), valid for \(x \geq 1\) where \(f\) is defined. The choice "\(\sqrt{x-1}+1\), for \(x \geq 1\)" is wrong because it fails to square the square root term correctly before adding 1, leaving the radical unsimplified. When composing functions involving roots and squares, carefully simplify the algebra step by step rather than leaving expressions partially combined.

The horizontal shift \(x - 2\) shifts every domain value to the right by 2, so the new domain becomes \([-3+2, 5+2] = [-1, 7]\); the vertical shift of \(+1\) does not affect domain. The choice "\([-3, 5]\)" is wrong because it ignores the horizontal transformation entirely, applying no shift when one clearly exists inside the function argument. Always apply horizontal transformations to the domain and vertical transformations to the range, keeping the two separate.

Since \((f \circ g)(x) = [g(x)]^2\), we need \([g(x)]^2 = 4x^2 - 12x + 9 = (2x-3)^2\), so \(g(x) = 2x - 3\) satisfies this exactly. The choice "\(g(x) = 2x+3\)" is wrong because squaring it gives \(4x^2 + 12x + 9\), which has the wrong sign on the middle term compared to the target expression. When reverse-engineering an inner function from a squared composition, factor the resulting quadratic as a perfect square to identify the correct linear expression.

Setting the two pieces equal at \(x = 1\) gives \(k(1)+1 = 3(1)-k\), which simplifies to \(k+1 = 3-k\), or \(2k = 2\), so \(k = 1\). The choice "\(k=2\)" is wrong because substituting it gives \(2(1)+1=3\) on the left but \(3(1)-2=1\) on the right, values that do not match. To find an unknown parameter that makes a piecewise function continuous, set the two pieces equal at the shared boundary and solve algebraically.

Completing the square gives \(f(x) = (x+1)^2 - 1\); setting \(y = (x+1)^2 -1\), solving for \(x\) gives \(x = -1 + \sqrt{y+1}\) (taking the positive root since \(x \geq -1\)), so \(f^{-1}(x) = -1+\sqrt{x+1}\). The choice "\(f^{-1}(x) = -1 - \sqrt{x+1}\)" is wrong because it selects the negative root, which would produce outputs less than \(-1\), contradicting the restricted domain used to define the inverse. When inverting a restricted quadratic function, always choose the root sign that matches the given domain restriction of the original function.

Since \(f(-x) = -f(x)\) and \(g(-x) = g(x)\), we get \(h(-x) = f(-x)g(-x) = -f(x)g(x) = -h(x)\), which is exactly the definition of an odd function. The choice "\(h\) is even" is wrong because it would require \(h(-x) = h(x)\), but the negative sign from the odd function carries through the product, preventing that equality. Memorize that the product of an odd and an even function is always odd, just as the product of two odds is even and two evens is even.

The vertical asymptote at \(x = 2\) excludes that value from the domain, and the horizontal asymptote created by the added constant 3 excludes \(y=3\) from the range since the fractional term never actually reaches zero. The choice "Domain: \(x \neq 3\); Range: \(y \neq 2\)" is wrong because it swaps the roles of the vertical and horizontal asymptotes, misplacing which value affects domain versus range. When transforming a basic rational function by shifting, remember that horizontal shifts affect the domain restriction while vertical shifts affect the range restriction.

We need the inner expression to satisfy \(0 \leq 2x-4 \leq 10\); solving gives \(4 \leq 2x \leq 14\), then dividing by 2 gives \(2 \leq x \leq 7\), so the domain is \([2,7]\). The choice "\([0,10]\)" is wrong because it applies the original domain directly to \(x\) without accounting for the horizontal compression and shift introduced by \(2x - 4\). When finding the domain of a transformed function, set the entire inner expression within the bounds of the original domain and solve for \(x\).

A graph is symmetric about the origin when \(f(-x) = -f(x)\), and testing \(f(-x) = (-x)^3-4(-x) = -x^3+4x = -(x^3-4x) = -f(x)\) confirms this property holds for this function since it contains only odd-degree terms. The choice "\(f(x) = x^2-4\)" is wrong because it only contains even-degree terms, making it symmetric about the \(y\)-axis instead of the origin. To identify origin symmetry, check that every term in the function has an odd exponent, with no even-degree or constant terms present.

Simplifying \(\frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3\) for \(x \neq 3\), the limit as \(x \to 3\) equals 6, but the function is defined to equal 5 there, creating a mismatch known as a removable discontinuity. The choice "\(f\) is continuous at \(x=3\) since the piece defines it there" is wrong because simply having a defined value does not guarantee continuity; the defined value must match the limit for continuity to hold. Always check that the limit of a function as \(x\) approaches a point equals the function's actual value there before declaring continuity.

Since \((g \circ f)(x) = (\sqrt{4-x})^2\), the domain is still restricted by the requirement that the inner radicand \(4-x\) be non-negative even though the square root and square operations appear to cancel algebraically, giving \(x \leq 4\). The choice "All real numbers" is wrong because it ignores that \(f(x)\) must be defined before \(g\) can be applied, so the domain restriction from the inner function carries through the composition regardless of algebraic simplification. Always determine the domain of a composed function based on the original inner function's restrictions, not just the simplified final expression.