Q. 42 - Calculus

Question

Use a sign chart for $f\text{'}\text{'}$to determine the intervals on which each function f in Exercises 41–52 is concave up or concave down, and identify the locations of any inflection points. Then verify your algebraic answers with graphs from a calculator or graphing utility

$f\left(x\right)=(x-3{)}^3(x-1)$

Step-by-Step Solution

Answer

The function is concave up on both $(-\infty ,2),(3,\infty )$ and concave down on $(2,3).$

Step 1. Given Information.

The given function is $f\left(x\right)=(x-3{)}^3(x-1)$

Step 2. Second derivative.

On differentiating, we get,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}(x-3{)}^3(x-1) \\ =\left(\frac{d}{dx}(x-3{)}^3\right)(x-1)+(x-3{)}^3\left(\frac{d}{dx}(x-1)\right) \\ =3(x-3{)}^2(x-1)+(x-3{)}^3 \\ =2(x-3{)}^2(2x-3\left) \\ {f}^{\text{'}\text{'}}\right(x)=\frac{d}{dx}2(x-3{)}^2(2x-3) \\ =2\left(\frac{d}{dx}(x-3{)}^2(2x-3)\right) \\ =2\left(2(x-3)(2x-3)+2(x-3{)}^2\right) \\ =12(x-3)(x-2) \end{gathered}$$

Step 3. Sign chart.

Now,

$f\text{'}\text{'}\left(x\right)=0 when x=2\text{ and }x=3,$.

Therefore, the sign chart will be,

The function has inflexion points $x=2,3$.

$\text{The function is concave up on both }(-\infty ,2),(3,\infty )\text{ and concave down on }(2,3)$

Step 4. Verification.

The graph of the function is,