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

Question

Isabella Ackerman · Accepted Answer

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

To determine if #f(x)# is increasing or decreasing at a specific #x#, first of all, you need to compute the derivative. Here, the quotient rule is applicable: 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# Your derivative is therefore: #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#.

Evelyn Agard · Answer

undefined To determine whether $ f(x) = \frac{x^2e^x}{x+2} $ is increasing or decreasing at $ x = -1 $, we need to evaluate the sign of the derivative at that point.

1. Find the derivative of $ f(x) $ using the quotient rule.
$$ f'(x) = \frac{(x+2)(2xe^x + x^2e^x) - (x^2e^x)(1)}{(x+2)^2} $$

2. Evaluate the derivative at $ x = -1 $.
$$ f'(-1) = \frac{(-1+2)(2(-1)e^{-1} + (-1)^2e^{-1}) - ((-1)^2e^{-1})(1)}{(-1+2)^2} $$

3. Simplify the expression.
$$ f'(-1) = \frac{(1)(-2e^{-1} + e^{-1}) - e^{-1}}{(1)^2} $$
$$ f'(-1) = -\frac{e^{-1}}{1} = -\frac{1}{e} $$

4. Since the derivative $ f'(-1) = -\frac{1}{e} $ is negative, $ f(x) $ is decreasing at $ x = -1 $.

Axel Alvarez · Answer

undefined To determine whether the function $ f(x) = \frac{x^2e^x}{x+2} $ is increasing or decreasing at $ x = -1 $, we can analyze the sign of the derivative at that point. Using the quotient rule and the product rule of differentiation, we find the derivative of $ f(x) $ to be:

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

Evaluating this derivative at $ x = -1 $ gives:

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

After simplifying this expression, we find:

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

Since the derivative $ f'(-1) = -2e^{-1} $ is negative, the function $ f(x) $ is decreasing at $ x = -1 $.