Review question

# For the functions $f$ and $g$, is $g(f(A))$ always bigger than $f(g(A))$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9099

## Solution

Let

$f(x) = \int_0^1(xt)^2\:dt, \quad \text{and} \quad g(x) = \int_0^xt^2\:dt.$

Let $A>0$. Which of the following statements is true?

1. $g(f(A))$ is always bigger than $f(g(A))$.

2. $f(g(A))$ is always bigger than $g(f(A))$.

3. They are always equal.

4. $f(g(A))$ is bigger if $A < 1$, and $g(f(A))$ is bigger if $A > 1$.

5. $g(f(A))$ is bigger if $A < 1$, and $f(g(A))$ is bigger if $A > 1$.

We have $f(x) = \int_0^1(xt)^2\:dt= x^2\int_0^1 t^2 \:dt = x^2\left[\dfrac{t^3}{3}\right]_0^1= \dfrac{x^2}{3}$ and $g(x) = \int_0^xt^2\:dt= \left[\dfrac{t^3}{3}\right]_0^x= \dfrac{x^3}{3}.$

Thus $f(g(A)) = f\left(\dfrac{A^3}{3}\right) = \dfrac{A^6}{3^3}$ and $g(f(A)) = g\left(\dfrac{A^2}{3}\right) = \dfrac{A^6}{3^4}.$

Now it is always true that $\dfrac{A^6}{3^3}>\dfrac{A^6}{3^4}$, and so (b) is our answer.