What is the slope of #f(x)=-e^x/(x-2# at #x=-2#?

Question

William Abdalla · Accepted Answer

#f'(- 2) = frac(5)(16 e^(2))#

We have: #f(x) = - frac(e^(x))(x - 2)#

First, let's find the function for the slope, #f'(x)#, by differentiating #f(x)#:

#Rightarrow f'(x) = frac(d)(dx)(- frac(e^(x))(x - 2))#

#Rightarrow f'(x) = - frac(d)(dx)(frac(e^(x))(x - 2))#

Using the quotient law of differentiation:

#Rightarrow f'(x) = - frac((x - 2) cdot frac(d)(dx)(e^(x)) - (e^(x)) cdot frac(d)(dx)(x - 2))((x - 2)^(2))#

#Rightarrow f'(x) = - frac((x - 2) cdot e^(x) - (e^(x)) cdot 1)((x - 2)^(2))#

#Rightarrow f'(x) = - frac(x e^(x) - 2 e^(x) - e^(x))((x - 2)^(2))#

#Rightarrow f'(x) = - frac(x e^(x) - 3 e^(x))((x - 2)^(2))#

#therefore f'(x) = - frac(e^(x) (x - 3))((x - 2)^(2))#

Now, we need to evaluate the slope at #x = - 2#.

So let's substitute #-2# in place of #x#:

#Rightarrow f'(- 2) = - frac(e^((- 2)) ((- 2) - 3))(((- 2) - 2)^(2))#

#Rightarrow f'(- 2) = - frac(e^(- 2) (- 5))((- 4)^(2))#

#Rightarrow f'(- 2) = - frac(- frac(5)(e^(2)))(16)#

#therefore f'(- 2) = frac(5)(16 e^(2))#

Therefore, at #x = - 2#, the slope of #f(x)# is #frac(5)(16 e^(2))#.

Scarlett Allison · Answer

undefined To find the slope of the function $f(x) = -\frac{e^x}{x - 2}$ at $x = -2$, you need to find the derivative of the function with respect to $x$ and then evaluate it at $x = -2$.

The derivative of $f(x)$ is found using the quotient rule:

$$f'(x) = \frac{(x - 2)(-e^x) - (-e^x)(1)}{(x - 2)^2}$$

Now, substitute $x = -2$ into the derivative:

$$f'(-2) = \frac{(-2 - 2)(-e^{-2}) - (-e^{-2})(1)}{(-2 - 2)^2}$$

Calculate the values:

$$f'(-2) = \frac{(-4)(-e^{-2}) + e^{-2}}{16}$$

$$f'(-2) = \frac{4e^{-2} + e^{-2}}{16}$$

$$f'(-2) = \frac{5e^{-2}}{16}$$

So, the slope of $f(x) = -\frac{e^x}{x - 2}$ at $x = -2$ is $ \frac{5e^{-2}}{16} $.