Home > Calculus > Interpreting the Sign of the First Derivative (Increasing and Decreasing Functions)

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

Answer 1

Ezekiel Alexander

decreasing.

You need to remember these rules:

So #f(x)=(x-2)e^x#; to differentiate we need to use the product rule: #d/dx(uv)=u(dv)/dx+(du)/dxv#

Applying this rule gives us:

# f'(x)=(x-2)(e^x)+(1)(e^x) # # :. f'(x)=xe^x-2e^x+e^x # # :. f'(x)=xe^x-e^x# # :. f'(x)=(x-1)e^x #

When # x=-2=>f'(-2)=(-2-1)e^-2 # When # :. f'(-2)=-3e^-2 #

Remember that # A^x > 0 AA x in RR => e^-2 > 0#

Hence, # f'(-2)=-3e^-2 < 0 # so we can conclude that when #x=-2# then #f(x)# is decreasing .

A graph always helps to visualise and confirm these solutions: graph{(x-2)e^x [-10.114, 7.666, -4.41, 4.48]}

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Answer 2

Ezekiel Alexander

To determine if ( f(x) = (x - 2)e^x ) is increasing or decreasing at ( x = -2 ), we need to evaluate the sign of the derivative at that point.

First, find the derivative of ( f(x) ): [ f'(x) = e^x(x - 1) ]

Now evaluate ( f'(-2) ): [ f'(-2) = e^{-2}(-2 - 1) = -\frac{3}{e^2} ]

Since ( f'(-2) ) is negative, ( f(x) ) is decreasing at ( x = -2 ).

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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.