Function Composition - IB Mathematics Analysis and Approaches

Let $f(x) = \frac{1}{x}$ and $g(x) = x-3$ . The domain of $g$ is restricted to $\{x \in \mathbb{R} | x > 5\}$ . Find the range of the composite function $(f \circ g)(x)$ .
Answer: $(0, 1/2)$ To find the range of $(f \circ g)(x)$ , we first determine the range of the inner function, $g(x)$ , over its given domain.\ The domain of $g(x)=x-3$ is $x > 5$ .\ If $x > 5$ , then $x-3 > 5-3$ , so $g(x) > 2$ .\ The range of $g(x)$ on this domain is $(2, \infty)$ .\ This range becomes the domain for the outer function, $f(x) = 1/x$ . We need to find the range of $f(u) = 1/u$ for $u \in (2, \infty)$ .\ The function $f(u)=1/u$ is a decreasing function for $u>0$ .\ As $u$ approaches 2 from the right ( $u \to 2^+$ ), $f(u) \to 1/2$ .\ As $u$ approaches infinity ( $u \to \infty$ ), $f(u) \to 0$ .\ Since the function is decreasing, it will take all values between these limits. Thus, the range of $(f \circ g)(x)$ is $(0, 1/2)$ .\ Distractor C is the range of $g(x)$ . Distractor D results from incorrectly evaluating the behavior of $1/u$ . Distractor B results from substituting the endpoint $x=5$ into $f(x)$ instead of $f(g(x))$ .
Let $f(x) = e^{x-1}$ and $g(x) = \ln(x) + 1$ . Find the value of $(g \circ f)(5) + (f \circ g)(e)$ .
Answer: $5+e$ A student might notice that $f(x)$ and $g(x)$ are inverse functions. To verify, find the inverse of $f(x)$ : let $y = e^{x-1}$ . Then $\ln(y) = x-1$ , so $x = \ln(y) + 1$ . Thus, $f^{-1}(x) = \ln(x) + 1 = g(x)$ .\ By the property of inverse functions, $(g \circ f)(x) = (f^{-1} \circ f)(x) = x$ and $(f \circ g)(x) = (f \circ f^{-1})(x) = x$ .\ Therefore, $(g \circ f)(5) = 5$ and $(f \circ g)(e) = e$ .\ The sum is $5 + e$ .\ Alternatively, one can compute each part directly:\ $f(5) = e^{5-1} = e^4$ .\ $(g \circ f)(5) = g(e^4) = \ln(e^4) + 1 = 4+1 = 5$ .\ $g(e) = \ln(e) + 1 = 1+1 = 2$ .\ $(f \circ g)(e) = f(2) = e^{2-1} = e^1 = e$ .\ The sum is $5+e$ .
Let $f(x) = 2x - 5$ and $g(x) = ax + b$ . The composite function $(f \circ g)(x)$ is a constant function with a value of 7 for all $x$ . Find the value of $b$ .
Answer: $6$ First, find the expression for the composite function $(f \circ g)(x)$ .\ $(f \circ g)(x) = f(g(x)) = f(ax+b) = 2(ax+b) - 5 = 2ax + 2b - 5$ .\ We are told this is a constant function. For a linear function $mx+c$ to be constant, the slope $m$ must be zero. In our expression $2ax + (2b-5)$ , the coefficient of $x$ is $2a$ .\ So, we must have $2a = 0$ , which implies $a=0$ .\ This means the function simplifies to $(f \circ g)(x) = 2(0)x + 2b - 5 = 2b - 5$ .\ We are given that the constant value is 7. Therefore:\ $2b - 5 = 7\ \(2b = 12\ \(b = 6$ .\ Distractor A is the value of $a$ . Distractor B is the value of $b$ if $2b-5=0$ . Distractor C is the value of $b$ if $2b=7$ .
Let $f(x) = |x - 2|$ and $g(x) = 3x + 1$ . Find the set of values of $x$ for which $(f \circ g)(x) < 5$ .
Answer: $-\frac{4}{3} < x < 2$ First, find the composite function $(f \circ g)(x)$ :\ $(f \circ g)(x) = f(g(x)) = f(3x+1) = |(3x+1) - 2| = |3x - 1|$ .\ Now, we solve the inequality $|3x - 1| < 5$ .\ This absolute value inequality is equivalent to the double inequality:\ $-5 < 3x - 1 < 5\ Add 1 to all parts:\ \(-4 < 3x < 6\ Divide by 3:\ \(-\frac{4}{3} < x < 2\ Distractor B comes from composing in the wrong order: \((g \circ f)(x) = g(|x-2|) = 3|x-2|+1$ . Solving $3|x-2|+1 < 5$ gives $|x-2| < 4/3$ , which leads to $2/3 < x < 10/3$ . Distractor C comes from misinterpreting the absolute value inequality as $>$ instead of $<$ . Distractor D comes from only solving one side of the inequality, $3x-1 < 5$ .
Let $f(x) = x^2 + 1$ and $g(x) = 2x - 1$ . Find the values of $x$ for which $(f \circ g)(x) = (g \circ f)(x)$ .
Answer: $x = 1 \pm \frac{\sqrt{2}}{2}$ First, find the expressions for the composite functions.\ $(f \circ g)(x) = f(g(x)) = f(2x - 1) = (2x - 1)^2 + 1 = (4x^2 - 4x + 1) + 1 = 4x^2 - 4x + 2\ \((g \circ f)(x) = g(f(x)) = g(x^2 + 1) = 2(x^2 + 1) - 1 = 2x^2 + 2 - 1 = 2x^2 + 1\ Now, set them equal to each other:\ \(4x^2 - 4x + 2 = 2x^2 + 1\ \(2x^2 - 4x + 1 = 0\ This is a quadratic equation. Since it does not factor easily, use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ :\ $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)}\ \(x = \frac{4 \pm \sqrt{16 - 8}}{4}\ \(x = \frac{4 \pm \sqrt{8}}{4}\ \(x = \frac{4 \pm 2\sqrt{2}}{4}\ \(x = 1 \pm \frac{2\sqrt{2}}{4}\ \(x = 1 \pm \frac{\sqrt{2}}{2}$ \ Distractor B comes from an error in simplifying $\frac{4 \pm 2\sqrt{2}}{4}$ to $1 \pm \sqrt{2}$ . Distractors C and D arise from calculation errors in the quadratic formula.
Given that $(f \circ g)(x) = 4x^2 - 12x + 9$ and $g(x) = 2x - 3$ , which of the following is a possible expression for $f(x)$ ?
Answer: $f(x) = x^2$ We have $(f \circ g)(x) = f(g(x)) = f(2x-3)$ .\ We are given $f(2x-3) = 4x^2 - 12x + 9$ .\ We can try to express the right-hand side in terms of $g(x) = 2x-3$ .\ Notice that $4x^2 - 12x + 9$ is a perfect square trinomial: $(2x-3)^2$ .\ So, we have $f(2x-3) = (2x-3)^2$ .\ Let $u = 2x-3$ . Then the equation becomes $f(u) = u^2$ .\ This means the function $f$ squares its input. Therefore, $f(x) = x^2$ .\ Distractor D implies that $f$ is the identity function, which is incorrect. Distractors B and C, if composed with $g(x)$ , would not produce the given expression.
Let $f(x)$ be an even function and $g(x)$ be an odd function. Which of the following composite functions is always an even function?
Answer: Both $(f \circ g)(x)$ and $(g \circ f)(x)$ Recall the definitions: $f$ is even if $f(-x) = f(x)$ for all $x$ . $g$ is odd if $g(-x) = -g(x)$ for all $x$ .\ Let's test $h(x) = (f \circ g)(x)$ :\ $h(-x) = f(g(-x))$ \ Since $g$ is odd, $g(-x) = -g(x)$ .\ $h(-x) = f(-g(x))$ \ Since $f$ is even, $f(-u) = f(u)$ . Let $u=g(x)$ .\ $h(-x) = f(g(x)) = h(x)$ .\ Therefore, $(f \circ g)(x)$ is an even function.\ Now let's test $k(x) = (g \circ f)(x)$ :\ $k(-x) = g(f(-x))$ \ Since $f$ is even, $f(-x) = f(x)$ .\ $k(-x) = g(f(x)) = k(x)$ .\ Therefore, $(g \circ f)(x)$ is also an even function.\ Since both are always even, the correct option is C.
Let $f(x) = \frac{1}{x-1}$ and $g(x) = \frac{x}{x+1}$ . Find a simplified expression for $(f \circ g)(x)$ .
Answer: $-x-1$ To find $(f \circ g)(x)$ , we substitute $g(x)$ into $f(x)$ :\ $(f \circ g)(x) = f(g(x)) = f\left(\frac{x}{x+1}\right) = \frac{1}{\frac{x}{x+1} - 1}\ To simplify the expression, we work with the denominator:\ \(\frac{x}{x+1} - 1 = \frac{x}{x+1} - \frac{x+1}{x+1} = \frac{x - (x+1)}{x+1} = \frac{x - x - 1}{x+1} = \frac{-1}{x+1}\ Now substitute this back into the main expression:\ \((f \circ g)(x) = \frac{1}{\frac{-1}{x+1}} = 1 \cdot \frac{x+1}{-1} = -(x+1) = -x-1$ .\ Distractor A (1/x) is the result of composing in the wrong order: $(g \circ f)(x) = g(\frac{1}{x-1}) = \frac{1/(x-1)}{1/(x-1) + 1} = \frac{1}{1 + (x-1)} = \frac{1}{x}$ . Distractors B and C are from algebraic errors during simplification.
Let $f(x) = 2x + 3$ and $g(x) = \begin{cases} x^2 & \text{if } x \ge 1 \\ 2x-1 & \text{if } x < 1 \end{cases}$ . Find the value of $(g \circ f)(-2)$ .
Answer: $-3$ To find $(g \circ f)(-2)$ , we first compute the inner function $f(-2)$ .\ $f(-2) = 2(-2) + 3 = -4 + 3 = -1$ .\ Now we compute $g(f(-2))$ , which is $g(-1)$ .\ To evaluate $g(-1)$ , we must use the rule for the piecewise function $g(x)$ . Since $-1 < 1$ , we use the second piece: $g(x) = 2x-1$ .\ $g(-1) = 2(-1) - 1 = -2 - 1 = -3$ .\ Therefore, $(g \circ f)(-2) = -3$ .\ Distractor C (1) arises from using the wrong piece of the definition for $g(x)$ , calculating $g(-1) = (-1)^2 = 1$ . Distractor A (-7) comes from composing in the wrong order: $g(-2) = 2(-2)-1=-5$ , and then $f(-5)=2(-5)+3 = -7$ . Distractor D (9) would come from $(f \circ g)(-2)$ , where $g(-2)=-5$ , but with an error $f(-5)=2(-5)+3 = -7$ . Or maybe $g(-2)=-3$ error, then $f(-3)=-3$ . No, wait. Maybe $g(-2)=(-2)^2=4$ , then $f(4)=11$ . A distractor of 1 is very plausible.
Let $f(x) = 2 - \frac{1}{x}$ . Find the value of $(f \circ f \circ f)(2)$ .
Answer: $5/4$ We need to compute the function composition iteratively.\ First, find $f(2)$ :\ $f(2) = 2 - \frac{1}{2} = \frac{3}{2}\ Next, find \((f \circ f)(2) = f(f(2)) = f(\frac{3}{2})\ \(f(\frac{3}{2}) = 2 - \frac{1}{3/2} = 2 - \frac{2}{3} = \frac{6-2}{3} = \frac{4}{3}\ Finally, find \((f \circ f \circ f)(2) = f(f(f(2))) = f(\frac{4}{3})\ \(f(\frac{4}{3}) = 2 - \frac{1}{4/3} = 2 - \frac{3}{4} = \frac{8-3}{4} = \frac{5}{4}\ Distractors A and B are the results of the first and second iterations, respectively. Distractor D would only be correct if \(f(x)=x$ or if the function value cycled back to the start.
The radius $r$ of a circular ripple in a pond, in meters, is given by the function $r(t) = 10\sqrt{t}$ , where $t$ is the time in seconds after a stone is dropped. The area $A$ of a circle is given by $A(r) = \pi r^2$ . Which function $(A \circ r)(t)$ correctly represents the area of the ripple as a function of time?
Answer: $100\pi t$ We need to find the composite function $(A \circ r)(t) = A(r(t))$ .\ This means we substitute the expression for $r(t)$ into the function $A(r)$ .\ $r(t) = 10\sqrt{t}\ \(A(r) = \pi r^2\ \((A \circ r)(t) = A(10\sqrt{t}) = \pi (10\sqrt{t})^2 = \pi (10^2 (\sqrt{t})^2) = \pi(100t) = 100\pi t$ .\ Distractor A incorrectly computes $\pi \cdot r(t)$ . Distractor C comes from not squaring the $\sqrt{t}$ term. Distractor D comes from squaring $t$ instead of $\sqrt{t}$ .
Let $f(x) = x^2+4$ and $g(x) = \sin(x)$ . Find the range of the composite function $(f \circ g)(x)$ .
Answer: $[4, 5]$ The composite function is $(f \circ g)(x) = f(g(x)) = f(\sin(x)) = (\sin(x))^2 + 4$ .\ To find the range of the composite function, we first consider the range of the inner function, $g(x) = \sin(x)$ . The range of $\sin(x)$ is $[-1, 1]$ .\ This range becomes the effective domain for the outer function $f(x)$ . We need to find the range of $f(u) = u^2+4$ for $u \in [-1, 1]$ .\ The function $f(u) = u^2+4$ is a parabola opening upwards with its vertex at $u=0$ .\ The minimum value on the interval $[-1, 1]$ occurs at the vertex, $u=0$ , where $f(0) = 0^2+4 = 4$ .\ The maximum value occurs at the endpoints of the interval, $u=-1$ and $u=1$ . $f(1) = 1^2+4 = 5$ and $f(-1) = (-1)^2+4 = 5$ .\ Therefore, the range of $(f \circ g)(x)$ is $[4, 5]$ .\ Distractor C is the range of $f(x)$ for all real $x$ . Distractor D results from only checking the endpoints $u=\pm 1$ and ignoring the minimum at the vertex. Distractor B may result from confusion about the range of $\sin^2(x)$ and combining it with the range of $f(x)$ .
Let $f(x)=x^2$ , $g(x)=x+2$ , and $h(x)=2x$ . Find the sum of the values of $x$ for which $(f \circ g)(x) = (h \circ f)(x)$ .
Answer: $4$ First, find the expressions for the two composite functions.\ $(f \circ g)(x) = f(g(x)) = f(x+2) = (x+2)^2 = x^2 + 4x + 4$ .\ $(h \circ f)(x) = h(f(x)) = h(x^2) = 2(x^2) = 2x^2$ .\ Now, set the two expressions equal to each other:\ $x^2 + 4x + 4 = 2x^2\ Rearrange to form a standard quadratic equation:\ \(0 = 2x^2 - x^2 - 4x - 4\ \(0 = x^2 - 4x - 4\ Let the roots of this quadratic equation be \(\alpha$ and $\beta$ . The sum of the roots of a quadratic equation $ax^2+bx+c=0$ is given by $-b/a$ .\ In this case, $a=1, b=-4, c=-4$ .\ The sum of the values of $x$ is $-(-4)/1 = 4$ .\ Alternatively, one could solve for $x$ using the quadratic formula, get $x = 2 \pm 2\sqrt{2}$ , and then sum the roots: $(2+2\sqrt{2}) + (2-2\sqrt{2}) = 4$ .\ Distractor A results from a sign error in the sum of roots formula (using $b/a$ ). Distractor B would be the sum if the equation was $x^2-4=0$ .
The function $f(x) = \frac{ax+3}{x-2}$ , where $a$ is a constant, is a self-inverse function. This means that $(f \circ f)(x) = x$ for all $x$ in the domain. Find the value of $a$ .
Answer: $2$ For a function to be self-inverse, we need $f(f(x)) = x$ . Let's compute $f(f(x))$ :\\ $f(f(x)) = f\left(\frac{ax+3}{x-2}\right) = \frac{a\left(\frac{ax+3}{x-2}\right)+3}{\left(\frac{ax+3}{x-2}\right)-2}$ \\Simplifying the numerator: $a\left(\frac{ax+3}{x-2}\right)+3 = \frac{a(ax+3)+3(x-2)}{x-2} = \frac{a^2x+3a+3x-6}{x-2}$ \\Simplifying the denominator: $\left(\frac{ax+3}{x-2}\right)-2 = \frac{ax+3-2(x-2)}{x-2} = \frac{ax+3-2x+4}{x-2} = \frac{(a-2)x+7}{x-2}$ \\So $f(f(x)) = \frac{a^2x+3a+3x-6}{(a-2)x+7}$ \\For this to equal $x$ , we need $\frac{a^2x+3a+3x-6}{(a-2)x+7} = x$ \\Cross-multiplying: $a^2x+3a+3x-6 = x((a-2)x+7) = (a-2)x^2+7x$ \\Rearranging: $(a-2)x^2+7x-a^2x-3a-3x+6 = 0$ \\ $(a-2)x^2+(4-a^2)x+(6-3a) = 0$ \\For this to hold for all $x$ , all coefficients must be zero:\\ $a-2 = 0 \Rightarrow a = 2$ \\We can verify: when $a=2$ , the coefficient of $x^2$ is $0$ , the coefficient of $x$ is $4-4=0$ , and the constant term is $6-6=0$ .
Let $f(x) = 3x^2 - 5$ and $g(x) = x+2$ . The function $h(x)$ is such that $(f \circ h)(x) = 3x^2 + 12x + 7$ . Find the function $h(x)$ .
Answer: $h(x) = x+2$ We are given $(f \circ h)(x) = f(h(x)) = 3x^2 + 12x + 7$ .\ From the definition of $f(x)$ , we have $f(h(x)) = 3(h(x))^2 - 5$ .\ Therefore, we can set the two expressions for $(f \circ h)(x)$ equal:\ $3(h(x))^2 - 5 = 3x^2 + 12x + 7\ \(3(h(x))^2 = 3x^2 + 12x + 12\ Divide by 3:\ \((h(x))^2 = x^2 + 4x + 4\ Recognize the right side as a perfect square:\ \((h(x))^2 = (x+2)^2\ Taking the square root of both sides gives \(h(x) = x+2$ or $h(x) = -(x+2)$ . Only $h(x) = x+2$ is listed as an option.\ Distractor B finds $(h(x))^2$ but not $h(x)$ . Distractor C comes from an algebraic error. Distractor D might arise from a sign error in factoring.
The function $h(x) = 3(x+2)^2 - 5$ can be written as a composition $(f \circ g)(x)$ . If $g(x) = x+2$ , what is $f(x)$ ?
Answer: $f(x) = 3x^2 - 5$ We are given $h(x) = (f \circ g)(x) = f(g(x))$ .\ We have $h(x) = 3(x+2)^2 - 5$ and $g(x) = x+2$ .\ Substituting $g(x)$ into the expression for $h(x)$ gives $h(x) = 3(g(x))^2 - 5$ .\ Since $h(x) = f(g(x))$ , we can see that $f(g(x)) = 3(g(x))^2 - 5$ .\ This means the rule for the function $f$ is to take its input, square it, multiply by 3, and then subtract 5. Therefore, $f(x) = 3x^2 - 5$ .\ Alternatively, let $u = g(x) = x+2$ . Then $x = u-2$ . Substitute this into the expression for $h(x)$ : $f(u) = 3((u-2)+2)^2 - 5 = 3u^2 - 5$ . So $f(x) = 3x^2 - 5$ .\ Distractor C results from incorrectly substituting $x = u+2$ instead of $x=u-2$ . Distractor A and D represent other algebraic mistakes.
Let $f(x) = \cos(x)$ and $g(x) = 2x + \frac{\pi}{2}$ . Find the exact value of $(f \circ g)(\frac{\pi}{3})$ .
Answer: $-\frac{\sqrt{3}}{2}$ We need to calculate $(f \circ g)(\frac{\pi}{3}) = f(g(\frac{\pi}{3}))$ .\ First, evaluate the inner function $g(\frac{\pi}{3})$ :\ $g(\frac{\pi}{3}) = 2(\frac{\pi}{3}) + \frac{\pi}{2} = \frac{2\pi}{3} + \frac{\pi}{2} = \frac{4\pi}{6} + \frac{3\pi}{6} = \frac{7\pi}{6}$ .\ Now, evaluate the outer function $f$ with this result:\ $f(\frac{7\pi}{6}) = \cos(\frac{7\pi}{6})$ .\ The angle $\frac{7\pi}{6}$ is in the third quadrant. The reference angle is $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$ .\ We know $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ .\ Since cosine is negative in the third quadrant, $\cos(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2}$ .\ Distractor D ignores the negative sign for cosine in the third quadrant. Distractor B is the value of $\sin(\frac{7\pi}{6})$ . Distractor C has both a sign error and uses the wrong trigonometric function.
Let $f(x) = x^3$ and $g(x) = x + k$ . Given that $(f \circ g)(2) = 27$ , find the value of $k$ .
Answer: $1$ First, evaluate the composite function at $x=2$ .\ $(f \circ g)(2) = f(g(2))$ .\ Find the value of the inner function $g(2)$ :\ $g(2) = 2 + k$ .\ Now substitute this result into $f(x)$ :\ $f(g(2)) = f(2+k) = (2+k)^3$ .\ We are given that this value is 27:\ $(2+k)^3 = 27\ Take the cube root of both sides:\ \(2+k = 3\ \(k = 3 - 2 = 1$ .\ Distractor D (19) comes from composing in the wrong order: $(g \circ f)(2) = g(f(2)) = g(2^3) = g(8) = 8+k$ . Setting $8+k = 27$ gives $k=19$ . Distractor A (-5) arises from incorrectly taking the cube root: $2+k=-3$ . Distractor C is a simple arithmetic error.

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