# Let f(x) be a function given by f(x) = x^5+10x^4-7x+12 a. Find all cut points of f''(x) b. Draw...

## Question:

Let {eq}f(x) {/eq} be a function given by {eq}f(x) = x^5+10x^4-7x+12 {/eq}

a. Find all cut points of {eq}f''(x) {/eq}

b. Draw the concavity chart,

c. Find the intervals on which {eq}f(x) {/eq} is concave upward,

d. Find the intervals on which {eq}f(x) {/eq} is concave downward,

e. Find all points of inflection in the format {eq}(x,y) {/eq}.

## Concavity:

We say that a function is concave up (down) on an interval {eq}I {/eq} if {eq}f'(x) {/eq} is increasing (decreasing) on {eq}I {/eq}. In practice, we use the following test.

If {eq}f''(x)>0\, (f''(x)<0) {/eq} on {eq}I {/eq}, then {eq}f(x) {/eq} is concave up (down) on {eq}I {/eq}.

An inflection point is a point where the concavity changes.

## Answer and Explanation:

Let {eq}f(x)=x^5+10x^4-7x+12 {/eq}. We then have

{eq}f'(x)=5x^4+40x^3-7\\ f''(x)=20x^3+120x^2=20x^2(x-6) {/eq}

a. Setting {eq}f''(x)=0 {/eq} we have

{eq}x=0,\, 6 {/eq}

and since {eq}f''(x) {/eq} is defined for all {eq}x {/eq}, we conclude that the cut points are {eq}x=0 {/eq} and {eq}x=6 {/eq}.

b. The concavity chart is shown below

Interval | negative infinity to 0 | 0 to 6 | 6 to infinity |

Sign of f(x) | negative | negative | positive |

c. This tells us that {eq}f(x) {/eq} is concave up on {eq}(6,0) {/eq}

d. and that {eq}f(x) {/eq} is concave down on {eq}(-\infty,0) {/eq} and on {eq}(0,6) {/eq}.

e. The only place where concavity changes is {eq}x=6 {/eq} giving a single inflection point of

{eq}(6,f(6))=(6,6^5+10(6^4)-7(6)+12)=(6,20,706) {/eq}

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