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

May 11, 2018

$- 3$

#### Explanation:

First of all, let's compute the derivative of $f \left(x\right)$, indicated as $f ' \left(x\right)$:

$f \left(x\right) = - {x}^{3} - 3 \setminus \implies f ' \left(x\right) = - 3 {x}^{2}$

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 $n {x}^{n - 1}$. So, the derivative of ${x}^{3}$ is $3 {x}^{3 - 1} = 3 {x}^{2}$, and since we had a minus sign in front of it, we will have to change signs: the derivative of $- {x}^{3}$ is $- 3 {x}^{3 - 1} = - 3 {x}^{2}$.

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 $f \left(x\right)$ associated a $y$ with every $x$, in the same way our derivative $f ' \left(x\right)$ associates, for every point $x$, the slope of the line tangent to the graph in the point $\left(x , f \left(x\right)\right)$

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

$f ' \left(- 1\right) = - 3 {\left(- 1\right)}^{2} = - 3 \setminus \cdot 1 = - 3$

May 11, 2018

$\text{slope } = - 3$

#### Explanation:

$\text{the slope is the value of } f ' \left(- 1\right)$

$\Rightarrow f ' \left(x\right) = - 3 {x}^{2}$

$\Rightarrow f ' \left(- 1\right) = - 3 {\left(- 1\right)}^{2} = - 3$