# Find the open intervals where the function f(x) = 3x^3 + 9x^2 + 175x - 6 is concave upward or...

## Question:

Find the open intervals where the function {eq}f(x) = 3x^3 + 9x^2 + 175x - 6 {/eq} is concave upward or concave downward. Find any inflection points.

## Concavity:

We recall that if {eq}f''(x)>0 {/eq} on {eq}I {/eq}, then {eq}f(x) {/eq} is concave up on {eq}I {/eq}.

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

An inflection point is any point in the domain of the curve where the concavity changes.

## Answer and Explanation:

Let {eq}f(x)=3x^3+9x^2+175x-6 {/eq}. We then have

{eq}f'(x)=9x^2+18x+175\\ f''(x)=18x+18 {/eq}

We can then see that {eq}f''(x)=0 {/eq} when {eq}x=-1 {/eq}.

It follows that {eq}f''(x)<0 {/eq} on {eq}(-\infty,-1) {/eq} and {eq}f''(x)>0 {/eq} on {eq}(-1,\infty) {/eq}.

So we have {eq}f(x) {/eq} is concave up on {eq}(-1,\infty) {/eq} and concave down on {eq}(-\infty,-1) {/eq}. We also conclude that {eq}f(x) {/eq} has an inflection point at {eq}x=-1 {/eq}.

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