How do you find the intervals of increasing and decreasing using the first derivative given #y=(x+4)/x^2#?

Question

Aurora Alvarado · Accepted Answer

#y(x)# is strictly decreasing in #(-oo,-8)# and reaches a local minimum in #x=8#, then it increases in # (-8,0)#, and it is decreasing again in #(0,+oo)#

Analyze the function's first derivative:

#dy/dx = d/dx((x+4)/x^2 ) = (x^2-2x(x+4))/x^4 = -(x+8)/x^3#

and resolve the disparity:

#dy/dx < 0#

When the numerator and denominator are both positive or both negative, the inequality is satisfied, so

#dy/dx < 0# for #x in (-oo,-8) uu (0,+oo)#

and:

#dy/dx > 0# for #x in (-8,0)#

while the only stationary point is for #x=-8#.

So #y(x)# is strictly decreasing in #(-oo,-8)# and reaches a local minimum in #x=8#, then it increases in # (-8,0)#, and it is decreasing again in #(0,+oo)#

graph{x^2 / (x+4) [-1.1181, -6.118, -0.86, 1.64]}

graph{[-9, -7, -0.08, -0.04]} graph{(x+4)/x^2

William Abdalla · Answer

undefined To find the intervals of increasing and decreasing for the function $ y = \frac{x+4}{x^2} $ using the first derivative, follow these steps: 1. Find the first derivative of the function. 2. Determine the critical points by setting the first derivative equal to zero and solving for $ x $. 3. Use the first derivative test to determine the intervals of increasing and decreasing. Here are the detailed steps: 1. Find the first derivative: $$ y' = \frac{d}{dx} \left( \frac{x+4}{x^2} \right) $$ $$ y' = \frac{d}{dx} \left( x^{-1} + 4x^{-2} \right) $$ $$ y' = -x^{-2} - 8x^{-3} $$ 2. Find the critical points by setting $ y' $ equal to zero and solving for $ x $: $$ -x^{-2} - 8x^{-3} = 0 $$ $$ -1 - 8x^{-1} = 0 $$ $$ 8x^{-1} = -1 $$ $$ x^{-1} = -\frac{1}{8} $$ $$ x = -8 $$ 3. Use the first derivative test to determine the intervals of increasing and decreasing: a. Test the interval $ (-\infty, -8) $ with a value such as $ x = -9 $: $$ y'(-9) = -(-9)^{-2} - 8(-9)^{-3} = -\frac{1}{81} - \frac{8}{729} < 0 $$ Since $ y'(-9) < 0 $, the function is decreasing on $ (-\infty, -8) $. b. Test the interval $ (-8, \infty) $ with a value such as $ x = -7 $: $$ y'(-7) = -(-7)^{-2} - 8(-7)^{-3} = -\frac{1}{49} - \frac{8}{343} > 0 $$ Since $ y'(-7) > 0 $, the function is increasing on $ (-8, \infty) $. Therefore, the function $ y = \frac{x+4}{x^2} $ is decreasing on the interval $ (-\infty, -8) $ and increasing on the interval $ (-8, \infty) $.