# Locate all critical points a)(t-5)^2*(t+3)^5 b)(2t-3)^3*(6-2t)^4

Luca B. | Certified Educator

calendarEducator since 2011

starTop subjects are Math, Science, and Business

You should remember that critical values represent the solutions to the following equaation such that:

`((t-5)^2*(t+3)^5)' = 0`

You need to use the product rule such that:

`((t-5)^2)'*(t+3)^5 + (t-5)^2*((t+3)^5)' = 0`

`2(t-5)(t+3)+(t-5)^2*(5(t+3)^4) = 0`

You need to factor out `(t-5)(t+3) ` such that:

`(t-5)(t+3)(2 + 5(t-5)(t+3)^3) = 0`

You need to solve the following equations such that:

`{(t-5=0),(t+3=0),(2+5(t-5)(t+3)^3=0):}` `=gt{(t=5),(t=-3),(2+5(t-5)(t+3)(t+3)^2=0):}`

`2 + 5(t^2-2t-15)(t^2+6t+9)=0`

`2+5(t^4+6t^3+9t^2-2t^3-12t^2-18t-15t^2-90t-135)=0`

`2+5t^4+30t^3+45t^2-10t^3-60t^2-90t-75t^2-450t-675 = 0`

`5t^4+20t^3-90t^2-540t-673 = 0`

The equation has two real roots `t=4.9`  and `t=-2.6`  and two complex roots.

Hence, evaluating the critical values of the function yields `t=-3, t=-2.6, t=4.9`  and `t=5` .

check Approved by eNotes Editorial