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

2 Answers

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lemjay | High School Teacher | (Level 3) Senior Educator

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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)]`



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


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

For the first factor,

`(t-2)^2 = 0`



For the second factor,




And the last factor,




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` ,



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


`t=2` ,



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

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rakesh05 | High School Teacher | (Level 1) Assistant Educator

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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.


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`



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.`