Q. 13 - Precalculus Enhanced with Graphing Utilities

Question

Approximate the local maximum values and local minimum values of $f\left(x\right)=x^3-5x+1$ on $(-4, 4)$. Determine where the function is increasing and where it is decreasing. 

Step-by-Step Solution

Answer

The graph of the function over the given interval is given as 

For the function, the local maximum value is $5.3$ and it occurs at $x=-1.3$.

The local minimum value is $-3.3$ and it occurs at $x=1.3$.

The graph is increasing over the interval $\left(-4,-1.3\right) \mathrm{and} \left(1.3,4\right)$and it is decreasing over the interval $\left(-1.3,1.3\right)$.

Step 1. Given information.

The given function is $f\left(x\right)=x^3-5x+1$.

Step 2. Graph the function.

Using a graphing utility the graph of the function$f\left(x\right)=x^3-5x+1$ over the interval $\left(-4,4\right)$ is given as

Step 3. Local maximum and minimum value.

The function has a local maximum at $\left(-1.3\right)$, since for all $x$ close to $-1.3$, we have $f\left(x\right)\le -1.3$. The local maximum value is $f(-1.3)=5.3$.

The function has a local minimum at $1.3$ and the local minimum value is $f(1.3)=-3.3$.

Step 4. Increasing and Decreasing interval.

From the graph, it can be seen that the graph is increasing in the interval $\left(-4,-1.3\right) \mathrm{and} \left(1.3,4\right)$.

And the graph is decreasing in the intervals $\left(-1.3,1.3\right)$.