#### To determine

**To find:** The local maximum and local minimum by drawing a graph of the given function and then estimate the inflection points by using the graph of f″.

#### Answer

The local minimum and local maximum values of the given function are 0.68 and 1.5 respectively.

The inflection points of the function are x≈−0.2, −1.2

#### Explanation

**You may know:**

Local Maxima and Local Minima:

If f is any real valued function and *c* be an interior point in the domain of f. Then,

(i) The point *c* is called a point of local maxima if there is an h>0 such that,

f(c)>f(x) for all x in (c−h,c+h)

(ii) The point *c* is called a point of local minima if there is an h>0 such that,

f(c)<f(x) for all x in (c−h,c+h)

Inflection Point:

The inflection points are those point at which the second derivative of the function is zero.

Given:

The exponential function, f(x)=ex3−x

**Calculation:**

The given function is f(x)=ex3−x.

Differentiate the given function with respect to *x*.

f′(x)=ddx(ex3−x)=ex3−x⋅ddx(x3−x)=ex3−x(3x2−1)

Find the critical point of the function by equating the above derivative to zero.

f′(x)=0ex3−x(3x2−1)=03x2−1=0 (∵ ex3−x≠0)x2=13

x=±13≈± 0.6

Find the intercepts of the graph of the function.

Substitute x=0 in f(x)=ex3−x to find the *y*-intercept.

y=f(0)=e0=1

So, the *y*-intercept is (0,1).

Also, there is no *x*-intercept.

It can be noted that when *x* tends to negative infinity, the *y-*value tends to 0.

So, the x-axis, that is, y=0 is the asymptote to the graph of the given function.

The graph of the given function is shown below:

From the above graph, it can be noted that the local maximum occurs at the point x≈−0.6 and local minimum occurs at the point x≈0.6.

The local minimum and local maximum values can be found as follows:

Local minimum=f(13)=e(13)3−13=e133−13

=e−233≈0.68

Local maximum=f(−13)=e(−13)3+13=e−133+13

=e233≈1.5

Calculation of Local Maximum and Local Minimum using Calculus:

Find the second derivative of the given function.

f″(x)=ddx[ex3−x(3x2−1)]=ddx(ex3−x)⋅(3x2−1)+ex3−x⋅ddx(3x2−1)=ex3−x(3x2−1)⋅(3x2−1)+ex3−x⋅(6x)

=ex3−x[(3x2−1)2+6x]=ex3−x(9x4−6x2+6x+1)

It can be observed that,

f″(−13)=−23e239 (<0) and f″(13)=23e−239 (>0)

This shows that the local maximum occurs at the point x=−13(≈−0.6) and local minimum occurs at the point x=13(≈0.6).

To find the inflection point, draw the graph of f″.

**Final statement:** the inflection points of the function are x≈−0.2, −1.2.