Question

Let f be the function given by f(x)=2x^3−4x^2−5. What is the value of f′(−1) ?

Answer 1

f'(x) = 6x^2-8x

f'(-1) = 6(-1)^2 -8(-1)

      = 6 +8

      = 14

Answer 2

Functions can be related by taking their derivatives.

The value of f'(-1) is 14

The function is given as:

$\mathbf{f(x) = 2x^3 - 4x^2 - 5}$

First, we calculate f'(x)

To calculate f'(x), we simply differentiate f(x) with respect to x.

So, we have:

$\mathbf{f'(x) = 6x^2 - 8x - 0}$

Simplify

$\mathbf{f'(x) = 6x^2 - 8x}$

Substitute -1 for x, to calculate f'(-1)

$\mathbf{f'(-1) = 6(-1)^2 - 8(-1)}$

$\mathbf{f'(-1) = 6 +8}$

$\mathbf{f'(-1) = 14}$

Hence, the value of f'(-1) is 14

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