# Locate all critical points of s(t)=(t-2)^3 (t+7)^5

lemjay | Certified Educator

`s(t)=(t-2)^3(t+7)^5`

To determine the critical points, take the derivative of the given function.To do so, apply the product rule which is `(u*v)'=v*u'+u*v'` .

So, let `u=(t-2)^3`    and        `v=(t+7)^5`

And, take the derivative of u and v to get u' and v'.

`u'=3(t-2)^2`         and       `v'= 5(t+7)^4`

Then, plug-in u, v , u' and v' to the product rule of derivative.

`s'(t)=(t+7)^5*3(t-2)^2 + (t-2)^3*5(t+7)^4`

Factor out the GCF to simplify.

`s'(t)= (t-2)^2(t+7)^4[3(t+7)+5(t-2)]`

`s'(t)=(t-2)^2(t+7)^4(3t+21+5t-10)`

`s'(t)=(t-2)^2(t+7)^4(8t+11)`

Now that the derivative of the function is in simplified form, set s'(t) equal to zero.

`0=(t-2)^2(t+7)^4(8t+11)`

Then, set each factor equal to zero and solve for t.

For the first factor,

`(t-2)^2 = 0`

`t-2=0`

`t=2`

For the second factor,

`(t+7)^4=0`

`t+7=0`

`t=-7`

And the last factor,

`8t+11=0`

`8t=-11`

`t=-11/8`

So, the critical numbers are t=-7, 11/8, 2.

Next, plug-in the values of t to the given function to get their corresponding value of s.

`t=-7` ,

`s(-7)=(-7-2)^3(-7+7)^5=0`

`t=-11/8=-1.375` , `s(-1.375)=(-1.375-2)^3(-1.375+7)^5= -216488.1`

`t=2` ,

`s(2)=(2-2)^3(2+7)^5=0`

Hence, the critical points are (-7,0) , (-1.375, -216488.1)  and (2,0).

rakesh05 | Certified Educator

Given expression is `s(t)=(t-2)^3(t+7)^5` .

The critical point for the function is the point "t" where the tangent of the function become parallel to the axis of t. i.e. where  ds/dt=0.

`d/dt(s(t))=d/dt[(t-2)^3(t+7)^5]`

or,       `d/dt(s(t))=(t+7)^5d/dt(t-2)^3+(t-2)^3d/dt(t+7)^5`

`=(t+7)^5 3(t-2)^2+(t-2)^3 5(t+7)^4`

`=(t-2)^2(t+7)^4[3(t+7)+5(t-2)]`

`=(t-2)^2(t+7)^4[8t+11]`

Now to find the critical points we take `d/dt(s(t))=0` .

i.e.     `(t-2)^2(t+7)^4[8t+11]=0`

Now,        `(t-2)=0rArrt=2`  ,

`(t+7)=0rArrt=-7` ,

`(8t+11)=0rArrt=-11/8`  .

So, the critical points of the given function are  `t=2,-7,-11/8.`