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

1 Answer
Lotusbluete
Jan 22, 2016

Thus, #f(x)# is decreasing at #x = -1#.

Explanation:

To determine if #f(x)# is increasing or decreasing at a specific #x#, first of all, you need to compute the derivative.

Here, you can use the quotient rule:

if #f(x) = (g(x))/(h(x))#, then the derivative can be computed as follows:

#f'(x) = (g'(x) h(x) - g(x) h'(x)) /(h^2(x))#

For you, #g(x) = x^2e^x# and #h(x) = x+2#.

The derivatives of #g(x)# and #h(x)# are:

#g'(x) = x^2e^x + 2xe^x# (with the product rule)

#h'(x) = 1#

Thus, your derivative is:

#f'(x) = ((x^2e^x + 2xe^x)(x+2) - x^2e^x * 1) / (x+2)^2#

# = (e^x(x^2 + 2x)(x+2) - e^x x^2) / (x+2)^2#

# = (e^x[(x^2 + 2x)(x+2) - x^2]) / (x+2)^2#

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

Now, let's evaluate the derivative at #x = -1#:

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

# = e^(-1)(-1 - 3 -4) #

# = -8 e^(-1)#

As #e^(-1) > 0#,

#f'(-1) = -8 e^(-1) < 0#.

Thus, #f(x)# is decreasing at #x = -1#.

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