Question

The function fgraphed below is defined by a polynomial expression of degree 4. Use the graph to solve the exercise.

(a) A function value $f\left(a\right)$is a local maximum value of f if $f\left(a\right)$is the _____the value of f on some open interval containing a. From the graph of f we see that there are two local maximum values of f : One local maximum is _____ , and it occurs when $x=2$; the other local maximum is_____ , and it occurs whenrole="math" localid="1646230261606" $x=\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}$.

(b) A function value $f\left(a\right)$is a local minimum value of f if $f\left(a\right)$is the _____ the value of fon some open interval containing a. From the graph of f we see that there is one local minimum values of f. The local minimum value is_____ , and and it occurs when $x=\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}$.

Short answer

1. Write$\text{'}a\text{'}$ by filling all the gaps given in step 1.
2. Write$\text{'}b\text{'}$ by filling all the gaps in step 1.

Step-by-step solution

Part a. Step 1. Filling all the gap.

(a) maximum $7,6,5$.

Part a. Step 2. Concept of local maximum.

The function value $f\left(a\right)$is a local maximum value of f if$f\left(a\right)\ge f\left(x\right)$ when x is a near a.

Part a. Step 3. Determining the local maximum of f.

The local maximum value of f is at$x=2$ and at $x=5$. So, local maximum value$f\left(2\right)=7$ and$f\left(5\right)=6$.

Part b. Step 1. Filling all the gap.

(a) minimum $2,4$.

Part b. Step 2. Concept of local maximum.

The function value $f\left(a\right)$ is a local minimum value of if$f\left(a\right)\le f\left(x\right)$ when x is a near a.

Part b. Step 3. Determining the local minimum of f.

The local minimum value of f is $x=4$ at and local maximum value $f\left(4\right)=2$.