How do you find the intervals of increasing and decreasing using the first derivative given #y=(x^5-5x)/5#?

Question

Isabella Ackerman · Accepted Answer

If #y = f(x)#, then:

#f# is increasing, when #x in (-infty, -1)uu(1, +infty)#

#f# is decreasing, when #x in (-1, 1)# Let #y=f(x) = (x^2 - 5x)/5#. Then #f# is differentiable, since we can break it up like #1/5x^5 - x#, which is a polynomial, and infinitely differentiable. Its first derivative is: #d/dx(1/5x^5 - x) = x^4 - 1#. We want to find the roots of this expression, or where the derivative is zero: #x^4-1 = 0 => (x^2 - 1)(x^2 + 1) = 0 => (x-1)(x+1)(x^2+1) = 0# We can see that for #x = -1# or #x = 1#, #dy/dx = 0#. #x^4 - 1# is a polynomial with two roots. Since the coefficient of #x^4# is #1 > 0#, it holds that: #x^4 - 1# is positive, when #x in (-infty, -1)uu(1, +infty)# #x^4 - 1# is negative, when #x in (-1, 1)# When #dy/dx > 0#, #f# is increasing. When #dy/dx < 0#, #f# is decreasing. Consequently, #f# is increasing, when #x in (-infty, -1)uu(1, +infty)# #f# is decreasing, when #x in (-1, 1)# (Since #f#'s domain is #RR#, we can safely extend our analysis to the entirety of #(-infty, +infty)#).

Aurora Alvarado · Answer

undefined To find the intervals of increasing and decreasing using the first derivative, follow these steps: 1. Find the first derivative of the given function $ y = \frac{x^5 - 5x}{5} $. 2. Set the first derivative equal to zero and solve for $ x $ to find critical points. 3. Determine the sign of the first derivative in the intervals determined by the critical points. - If the first derivative is positive, the function is increasing in that interval. - If the first derivative is negative, the function is decreasing in that interval. Let's go through the steps: 1. Find the first derivative of $ y = \frac{x^5 - 5x}{5} $: $$ y' = \frac{d}{dx} \left( \frac{x^5 - 5x}{5} \right) $$ $$ y' = \frac{1}{5} \cdot \frac{d}{dx} (x^5 - 5x) $$ $$ y' = \frac{1}{5} (5x^4 - 5) $$ $$ y' = x^4 - 1 $$ 2. Set the first derivative equal to zero and solve for $ x $: $$ x^4 - 1 = 0 $$ $$ x^4 = 1 $$ $$ x = \pm 1 $$ 3. Determine the intervals of increasing and decreasing: - For $ x < -1 $, $ y' < 0 $, so the function is decreasing. - For $ -1 < x < 1 $, $ y' > 0 $, so the function is increasing. - For $ x > 1 $, $ y' > 0 $, so the function is increasing. Therefore, the intervals of increasing are $ (-\infty, -1) $ and $ (1, \infty) $, and the interval of decreasing is $ (-1, 1) $.