Question

Let $f$ be the function given by $f\left(x\right)=x^3-6x^2-15x$. What is the absolute maximum value of $f$ on the interval $\left[0,6\right]$?

• A. $0$
• B. $4$
• C. $5$
• D. $6$
• E. $8$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Step-1: Find the critical point:

Since $f\left(x\right)=x^3-6x^2-15x$ is a polynomial, it is continuous and differentiable everywhere. To find the absolute maximum of $f$ on the interval $\left[0,6\right]$, determine all its critical points in $\left(0,6\right)$ where its derivative $f\text{'}\left(x\right)$ vanishes.

$\begin{array}{cc}& \begin{array}{rcl}f\text{'}\left(x\right)& =& 0\end{array}\\ \Rightarrow & \frac{d\left(x^3-6x^2-15x\right)}{dx}=0\\ \Rightarrow & 3x^2-6\times \left(2x\right)-15=0\\ \Rightarrow & 3\left(x^2-2\times \left(2x\right)-5\right)=0\\ \Rightarrow & 3\left(x^2-4x-5\right)=0\\ \Rightarrow & x^2-4x-5=\frac{0}{3}\\ \Rightarrow & x^2-5x+x-5=0\\ \Rightarrow & x\left(x-5\right)+1\left(x-5\right)=0\\ \Rightarrow & \left(x-5\right)\left(x+1\right)=0\\ \Rightarrow & x=5,-1\end{array}$

Note that $-1\notin \left(0,6\right)$. Thus, $f\text{'}\left(x\right)$ vanishes only at $x=5$ in the interval $\left(0,6\right)$.

Step-2: Check for global maximum value:

Determine the value of $f\left(x\right)$ at the only critical point $x=5$, and the boundaries of the interval $\left[0,6\right]$, namely $x=0$, and $x=6$. The maximum value of $f\left(x\right)$ among these values is the absolute maximum value of $f$ on the interval $\left[0,6\right]$.

$x$	$f\left(x\right)=x^3-6x^2-15x$
$x=0$	$f\left(0\right)={\left(0\right)}^3-6{\left(0\right)}^2-15\left(0\right)=0$
$x=5$	$f\left(5\right)={\left(5\right)}^3-6{\left(5\right)}^2-15\left(5\right)=125-6\times 25-75=50-150=-100$
$x=6$	$f\left(6\right)={\left(6\right)}^3-6{\left(6\right)}^2-15\left(6\right)=216-6\times 36-90=216-216-90=-90$

Thus, the absolute maximum value of $f\left(x\right)$ on the interval $\left[0,6\right]$ is $\mathbf{0}$ when $x=0$.

Hence, option(A) is correct.