Question

Bounding the second derivative: Consider the function $f\left(x\right)=x^2(x-3)$.

(a) To the nearest tenths place, find the smallest number $M$so that $\left|f\text{'}\text{'}\left(x\right)\right|\le M$for all $x\in [0,2]$.

(b) To the nearest tenths place, find the smallest number $M$so that $\left|f\text{'}\text{'}\left(x\right)\right|\le M$for all $x\in [0,4]$.

Short answer

Part (a) The smallest value of$M$is$6$for all values $x\in [0,2]$.

Part (b) The smallest value of$M$is$6$for all values $x\in [0,4]$.

Step-by-step solution

Part (a) Step 1: Given Information

Consider the function.

$f\left(x\right)=x^2(x-3)$

Part (a) Step 2: Simplify the given function and then determine the second derivative.

Simplify the given function.

$\begin{array}{rcl}f\left(x\right)& =& x^2(x-3)\\ & =& x^3-3x^2\end{array}$

The first derivative is as follows:

$\begin{array}{rcl}f\text{'}\left(x\right)& =& 3x^2-6x\end{array}$

The second derivative is as follows:

$\begin{array}{rcl}f\text{'}\text{'}\left(x\right)& =& 6x-6\end{array}$

Part (a) Step 3: Determine the value of M such that f''(x)≤M for all x∈[0, 2].

$\begin{array}{rcl}\left|f\text{'}\text{'}\left(x\right)\right|& =& \left|6x-6\right|\\ & \le & \left|6x\right|+\left|6\right|\\ & \le & 6\left|x\right|+6\dots \left(1\right)\end{array}$

The expression $\left(1\right)$is the smallest when $\left|x\right|$is the smallest, which occurs when $\left|x\right|=0$for all$x\in [0,2]$.

$\begin{array}{rcl}\left|f\text{'}\text{'}\left(x\right)\right|& \le & 6\left(0\right)+6\\ & \le & 6\end{array}$

The value of $M$is $6$for all values $x\in [0,2]$.

Part (b) Step 1: Determine the value of M such that f''(x)≤M for all x∈[0, 4].

Consider the function$f\left(x\right)=x^2(x-3)$.

Thus, the expression $\left(1\right)$is the smallest $\left|x\right|$when is the smallest, which occurs when $\left|x\right|=0$for all $x\in [0,4]$.

The second

$\begin{array}{rcl}\left|f\text{'}\text{'}\left(x\right)\right|& \le & 6\left(0\right)+6\\ & \le & 6\end{array}$

Hence, the value of $M$is $6$for all values $x\in [0,4]$.