`f(t) = tsqrt(4 - t^2), [-1, 2]` Find the absolute maximum and minimum values of f on the given interval

Textbook Question

Chapter 4, 4.1 - Problem 55 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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mathace | (Level 3) Assistant Educator

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Given: `f(t)=tsqrt(4-t^2),[-1,2].`

Find the critical number by setting the first derivative equal to zero and solving for the t values. Find the derivative using the product rule.

`f'(t)=t[-t/(4-t^2)^(1/2)]+(4-t^2)^(1/2)=0`

`t^2/(4-t^2)^(1/2)=(4-t^2)^(1/2)`

`t^2=4-t^2`

`2t^2=4`

`t^2=2`

`t=+-sqrt(2)` 

The critical numbers are `x=+-sqrt(2).`

Substitute the critical numbers and the endpoints of the interval [-1,2] into the original f(t) function. Do NOT substitute in the `t=-sqrt(2)`  because it is not in the interval [-1,2].

`f(-1)=-sqrt(3)`

`f(sqrt(2))=2`

`f(2)=0`

Examine the f(x) values to determine the absolute maximum and absolute minimum.

The absolute maximum occurs at the point (`sqrt(2),2)` .

The absolute minimum occurs at the point `(-1,-sqrt(3)),`

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gsarora17 | (Level 2) Associate Educator

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`f(t)=tsqrt(4-t^2)`

differentiating by applying product rule,

`f'(t)=t(1/2)(4-t^2)^(-1/2)(-2t)+sqrt(4-t^2)`

`f'(t)=(-t^2)/sqrt(4-t^2)+sqrt(4-t^2)`

`f'(t)=(-t^2+4-t^2)/sqrt(4-t^2)`

`f'(t)=(-2t^2+4)/sqrt(4-t^2)`

Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.

Now to find the critical numbers, solve for x for f'(x)=0.

`(-2t^2+4)/sqrt(4-t^2)=0`

`-2t^2+4=0`

`2t^2=4`

`t=sqrt(2) , -sqrt(2)`

Now the critical number -`sqrt(2)` is not in the interval (-1,2), so we have to evaluate the function at the critical point `sqrt(2)` and at the end points of the interval.

`f(-1)=-1sqrt(4-(-1)^2)=-sqrt(3)`

`f(2)=2sqrt(4-2^2)=0`

`f(sqrt(2))=sqrt(2)sqrt(4-2)=2`

So the function has absolute maximum=2 at t=`sqrt(2)`

It has no absolute minimum which is clear from the graph.

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gsarora17 | (Level 2) Associate Educator

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Pl. note correction in absolute minimum.

Function has absolute minimum=- `sqrt(3)` at t=-1 

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