# f(x) = 3x^3 - 2x^2 + 15 find f''(1)

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To evaluate the value the second derivative, in a given point, we'll have to calculate the first derivative:

f'(x) = (3x^3 - 2x^2 + 15)'

f'(x) = 9x^2 - 4x + 0

f'(x) = 9x^2 - 4x

Now, we'll differentiate f'(x):

[f'(x)]' = f"(x) = (9x^2 - 4x)'

f"(x) = 18x - 4

Now, we can calculate f"(1), substituting x by 1 in the expression of f"(x):

f"(1) = 18*1 - 4

f"(1) = 18 - 4

**f"(1) = 14**

f(x) = 3x^3 - 2x^2 + 15

First let us calcuate the first drivative f'(x):

f'(x) = 3*3x^2 - 2*2x + 0

= 9x^2 - 4x

Now differentitae f'(x) so we can determine f''(x):

f''(x) = 2*9x - 4

= 18x - 4

==> f''(x) = 18x-4

Now substitute with x= 1:

==> f''(1) = 18*1 - 4

= 18-4 = 14

**==> f''(1) = 14**

f(x) = 3x^3-2x^2+15.

To find f"(1).

Differentiating the given function we get:

f'(x) = {3x^3-2x^2+15}'

f'(x) = (3x^3)'-(2x^2)' +(15)' as (u(x) +or- v(x))' = u'(x) + or - v'(x).

f'(x) = 3*3x^(3-1) - 2*2x^(2-1) +0 , as (kx^n)' = k*nx^(n-1), where k is a constant.

f'(x) = 9x^2 - 4x

f"(x) = (f'(x))' = (9x^2-4x)' = 9*2x -4 = 18x-4

f"(x) = 18x-4

f"(1) = 18*1 -4 = 14

f"(1) = 14.