Q 58. - Precalculus Enhanced with Graphing Utilities

Question

In Problems 53– 60, use a graphing utility to graph each function over the indicated interval and approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places.

$f\left(x\right)=-0.4x^3+0.6x^2+3x-2 (-4,5)$

Step-by-Step Solution

Answer

The required graph is shown below:

The local maximum is $3.25$ and local minimum is $-4.05$.

The function is increasing on the interval $(-1.16,2.16)$.

The function is decreasing on the intervals $(-4,-1.16)$ and $(3.25,5)$.

Step 1. Write the given function and draw the graph using a graphing utility.

The given function is:

$f\left(x\right)=-0.4x^3+0.6x^2+3x-2 (-4,5)$

The required graph is shown below:

Step 2. Determine the local maximum and local minimum.

A function $f$ has a local maximum at c if there is an open interval I  containing c  so that for all x in I , $f\left(x\right)\le f\left(c\right)$. We will call $f\left(c\right)$ a local maximum value of $f$.

A function $f$ has a local minimum at c if there is an open interval I  containing c  so that, for all x in I , $f\left(x\right)\ge f\left(c\right)$ . We call $f\left(c\right)$ a local minimum value of $f$.

From the graph that we have drawn, we can see that $f$ has local maximum at point $(2.16,3.25)$.

Here $f\left(x\right)<f(2.16)$.

Therefore, the local maximum is $f(2.16)=3.25$.

From the graph that we have drawn, we can see that $f$ has a local minimum at $(-1.16,-4.05)$.

Here $f\left(x\right)>f(-1.16)$.

Therefore, the local minimum is $f(-1.16)=-4.05$.

Step 3. Determine where the function is increasing and decreasing.

We can see from the graph that the function is increasing from the point $(-1.16,-4.05)$ to $(2.16,3.25)$.

So we can conclude from it that is increasing on the interval $(-1.16,2.16)$.

We can see from the graph that the function is decreasing on the intervals $(-4,-1.16)$ and $(3.25,5)$.