Q. 83 - Calculus

Question

In Exercises 83–86, use the given derivative $f\text{'}$ to find any local extrema and inflection points of f and sketch a possible graph without first finding a formula for f.

$f\text{'}\left(x\right)=x^3-3x^2+3x$

Step-by-Step Solution

Answer

The possible graph of f is

Step 1. Given Information.

The given derivative function is $f\text{'}\left(x\right)=x^3-3x^2+3x.$

Step 2. Apply the first derivative test.

To find the local extrema the first derivative of the function must be zero.

So, 

$$\begin{gathered} f\text{'}\left(x\right)=x^3-3x^2+3x \\ 0=x^3-3x^2+3x \\ 0=x\left(x^2-3x+3\right) \\ x=0 \end{gathered}$$

Thus, by the first derivative test, f has a local minimum at $x=0.$The function has no local maxima.

Step 3. Finding inflection points.

An inflection point occurs when $f\text{'}\text{'}\left(x\right)=0.$

To find the inflection points, use the second derivative test.

So,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=3x^2-6x+3 \\ 0=3x^2-6x+3 \\ 0=3\left(x^2-2x+1\right) \\ 0=\left(x-1\right)\left(3x-3\right) \\ x=1 \end{gathered}$$

Thus, the function has an inflection point at $x=1.$ It is positive everywhere. Hence the graph of f  will be concave up everywhere.

Step 4. Sketch the graph of function f.

The graph of the function is