Question

How do you find the absolute maximum and absolute minimum values of f on the given interval: `f(t) =t sqrt(25-t^2)` on [-1, 5]?

Answer 1

Reqd. extreme values are `-25/2 and 25/2`.

Explanation:

We use substitution `t=5sinx, t in [-1,5]`.

Observe that this substitution is permissible, because,

`t in [-1,5] rArr -1<=t<=5rArr -1<=5sinx<=5`

`rArr -1/5<=sinx<=1`,

which holds good, as range of `sin` fun. is `[-1,1]`.

Now, `f(t)=tsqrt(25-t^2)=5sinx*sqrt(25-25sin^2x)`

`=5sinx*5cosx=25sinxcosx=25/2(2sinxcosx)=25/2sin2x`

Since, `-1<=sin2x<=1 rArr -25/2<=25/2sin2x<=25/2`

`rArr -25/2<=f(t)<=25/2`

Therefore, reqd. extremities are `-25/2 and 25/2`.

Answer 2

Find the monotony of the function from the derivative's sign and decide which local maximum/minimums are the biggest, smallest.

Absolute maximum is:
`f(3.536)=12.5`

Absolute minimum is:
`f(-1)=-4.899`

Explanation:

`f(t)=tsqrt(25-t^2)`

The derivative of the function:

`f'(t)=sqrt(25-t^2)+t*1/(2sqrt(25-t^2))(25-t^2)'`

`f'(t)=sqrt(25-t^2)+t*1/(2sqrt(25-t^2))(-2t)`

`f'(t)=sqrt(25-t^2)-t^2/sqrt(25-t^2)`

`f'(t)=sqrt(25-t^2)^2/sqrt(25-t^2)-t^2/sqrt(25-t^2)`

`f'(t)=(25-t^2-t^2)/sqrt(25-t^2)`

`f'(t)=(25-2t^2)/sqrt(25-t^2)`

`f'(t)=2(12.5-t^2)/sqrt(25-t^2)`

`f'(t)=2(sqrt(12.5)^2-t^2)/sqrt(25-t^2)`

`f'(t)=2((sqrt(12.5)-t)(sqrt(12.5)+t))/sqrt(25-t^2)`

• The numerator has two solutions:
  `t_1=sqrt(12.5)=3.536`
  `t_2=-sqrt(12.5)=-3.536`
  Therefore, the numerator is:
  Negative for `t in(-oo,-3.536)uu(3.536,+oo)`
  Positive for `t in(-3.536,3.536)`

• The denominator is always positive in `RR`, since it's a square root.
  Finally, the range given is `[-1,5]`

Therefore, the derivative of the function is:
- Negative for `t in[-1,3.536)`
- Positive for `t in(3.536,5)`
This means the graph firstly goes up from `f(-1)` to `f(3.536)` and then goes down to `f(5)`. This makes `f(3.536)` the absolute maximum and the biggest value of `f(-1)` and `f(5)` is the absolute minimum.

Absolute maximum is `f(3.536)`:
`f(3.536)=3.536sqrt(25-3.536^2)=12.5`

For the absolute maximum:

`f(-1)=-1sqrt(25-(-1)^2)=-4.899`

`f(5)=5sqrt(25-5^2)=0`

Therefore, `f(-1)=-4.899` is the absolute minimum.

You can see from the graph below that this is true. Just ignore the area left of `-1` since it's out of the domain:

graph{xsqrt(25-x^2) [-14.4, 21.63, -5.14, 12.87]}