Is #f(x)=(x^2-e^x)/(x-2)# increasing or decreasing at #x=-1#?

Answer 1

Charles Anctil

Increasing.

If #f'(-1)<0#, then #f(x)# is decreasing at #x=-1#. If #f'(-1)>0#, then #f(x)# is increasing at #x=-1#.

Determine the initial derivative.

Apply the rule of quotient.

#f'(x)=((x-2)d/dx[x^2-e^x]-(x^2-e^x)d/dx[x-2])/(x-2)^2#

Determine each derivative on its own.

#d/dx[x^2-e^x]=2x-e^x#

#d/dx[x-2]=1#

Re-plug them in.

#f'(x)=((x-2)(2x-e^x)-(x^2-e^x))/(x-2)^2#

#f'(x)=(2x^2-4x-xe^x+2e^x-x^2+e^x)/(x-2)^2#

#f'(x)=(x^2-xe^x+3e^x-4x)/(x-2)^2#

Find #f'(-1)#.

#f'(-1)=((-1)^2-(-1)e^(-1)+3e^(-1)-4(-1))/((-1)-2)^2#

#f'(-1)=(5+4/e)/9#

We could determine the exact value of this number, but it's clear that it will be positive. Because of this, we know that #f(x)# is increasing when #x=-1#.

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Answer 2

Isabella Ackerman

To determine if $f(x) = \frac{x^2 - e^x}{x - 2}$ is increasing or decreasing at $x = -1$ , evaluate the derivative of $f(x)$ at $x = -1$ and check its sign.

The derivative of $f(x)$ with respect to $x$ is $f'(x) = \frac{(2x - e^x)(x - 2) - (x^2 - e^x)}{(x - 2)^2}$ .

Evaluate $f'(-1)$ to find if $f(x)$ is increasing or decreasing at $x = -1$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.