Given that f(x)=x^6h(x)
h(-1)=3
h'(-1)=6
Calculate f'(-1).
f'= 6x^5 * h + h'*x^6
you know x, h(x), and h'(x). compute.
I'm lost how how they provide h(-1)=3 and h'(-1)=6
Would x=-1 and the h be 3, and h' be 6 in 6x^5 * h + h'*x^6 ? Or would the h be -1 and the h' be -1 and then I put the 3 and 6 outside the equation like 6x^5 * h + h'*x^6 =3+6?
If h(-1) = 3 then f(-1) = (-1)^6 * h(-1) = (-1)^6 * 3
To calculate f'(-1), we need to find the derivative of f(x) with respect to x and then substitute x = -1 into the resulting expression.
Let's start by applying the product rule to find the derivative of f(x) = x^6h(x):
f'(x) = (x^6)' * h(x) + x^6 * h'(x)
The derivative of x^6 can be determined by applying the power rule:
(x^6)' = 6x^(6-1) = 6x^5
Now, we'll use the given information to substitute values:
f'(-1) = (6(-1)^5) * h(-1) + (-1)^6 * h'(-1)
Since h(-1) = 3 and h'(-1) = 6:
f'(-1) = (6(-1)^5) * 3 + (-1)^6 * 6
Now, simplify the expression:
f'(-1) = (-6) * 3 + 1 * 6
f'(-1) = -18 + 6
f'(-1) = -12
Therefore, f'(-1) = -12.