How do you find the intervals of increasing and decreasing using the first derivative given #y=x^2-6x+8#?

Question

Ezekiel Alexander · Accepted Answer

See below

#y'=2x-6# #y'=0# at #x=3# For #x < 3#, #y'# is negative (for example at #x=1#, #y'=-4#) so #y# is decreasing on the interval #(-oo,3)# For #x > 3#, #y'# is positive (for example at #x=4#, #y'=2#) so #y# is increasing on the interval #(3,oo)#

Ezra Adams · Answer

undefined To find the intervals of increasing and decreasing for the function $ y = x^2 - 6x + 8 $ using the first derivative, follow these steps:

1. Find the first derivative of the function $ y' $.
2. Set $ y' $ equal to zero and solve for $ x $ to find critical points.
3. Test the intervals between critical points by plugging test points into $ y' $ to determine if the function is increasing or decreasing in those intervals.

Here are the detailed steps:

1. The first derivative of $ y = x^2 - 6x + 8 $ is $ y' = 2x - 6 $.
2. Set $ y' = 0 $ and solve for $ x $:
$$ 2x - 6 = 0 $$
$$ 2x = 6 $$
$$ x = 3 $$

So, the critical point is $ x = 3 $.
3. Test the intervals:
   - Choose a test point less than 3, say $ x = 0 $.
     $$ y'(0) = 2(0) - 6 = -6 $$
     Since $ y'(0) < 0 $, the function is decreasing on the interval $ (-\infty, 3) $.
   - Choose a test point greater than 3, say $ x = 4 $.
     $$ y'(4) = 2(4) - 6 = 2 $$
     Since $ y'(4) > 0 $, the function is increasing on the interval $ (3, \infty) $.

Therefore, the interval of increasing is $ (3, \infty) $, and the interval of decreasing is $ (-\infty, 3) $.