Question

What is the slope of `f(x)=-x^3-3` at `x=-1`?

Answer 1

`-3`

Explanation:

First of all, let's compute the derivative of `f(x)`, indicated as `f'(x)`:

`f(x) = -x^3-3 \implies f'(x) = -3x^2`

In fact, to derive a sum you must derive each single term. The first term is a power of `x`, and the derivative of `x^n` is `nx^{n-1}`. So, the derivative of `x^3` is `3x^{3-1} = 3x^2`, and since we had a minus sign in front of it, we will have to change signs: the derivative of `-x^3` is `-3x^{3-1} = -3x^2`.

As for the second term, the derivative of a number is always zero, which is why the term `-3` has disappeared in the derivative.

Now, just like our function `f(x)` associated a `y` with every `x`, in the same way our derivative `f'(x)` associates, for every point `x`, the slope of the line tangent to the graph in the point `(x,f(x))`

In other words, `f'(-1)` is exactly the answer you are looking for. The computation is

`f'(-1) = -3(-1)^2 = -3\cdot 1 = -3`

Answer 2

`"slope "=-3`

Explanation:

`"the slope is the value of "f'(-1)`

`rArrf'(x)=-3x^2`

`rArrf'(-1)=-3(-1)^2=-3`