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borishaifa [10]
3 years ago
10

Pictures attached.

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
3 0

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


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