Question

Pictures attached.

I'm like 99.9% sure that my inflection point is correct!??

Also, nothing i try for the local extremes is working...

Answer 1

we are given

$f(x)=4x^3+21x^2-294x+7$

For finding inflection point, we will find second derivative

$f'(x)=4\times 3x^2+21\times 2x-294+0$

$f'(x)=12x^2+42x-294$

now, we can find derivative again

$f''(x)=12\times 2x+42-0$

$f''(x)=24x+42$

now, we can set it to 0

and then we can solve for x

$f''(x)=24x+42=0$

we get

$x=-\frac{7}{4}$

we know that inflection point is a point where concavity changes

so, we will draw a number line and locate x=-7/4

and then we can find sign of second derivative on each interval

Concave up interval:

$(-\frac{7}{4},\infty)$

Concave down interval:

$(-\infty,-\frac{7}{4})$

Since, concavity changes at x=-7/4

So, there will be inflection point at x=-7/4

now, we can find y-value

$f(-\frac{7}{4}) =4(-\frac{7}{4})^3+21(-\frac{7}{4})^2-294(-\frac{7}{4})+7$

$f(-\frac{7}{4}) =\frac{4515}{8}$

So, the inflection point is

$(-\frac{7}{4},\frac{4515}{8})$..............Answer