Question

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]

Answer 1

Answer: Absolute minimum: f(-1) = -2$\sqrt{6}$

              Absolute maximum: f($\sqrt{12.5}$) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = $t\sqrt{25-t^{2}}$

f'(t) = $1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)$

f'(t) = $\sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }$

f'(t) = $\frac{25-2t^{2}}{\sqrt{25-t^{2}} }$ = 0

For this function to be zero, only denominator must be zero:

$25-2t^{2} = 0$

t = ±$\sqrt{2.5}$

$\sqrt{25-t^{2}}$ ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = $-\sqrt{2.5}$ and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = $t\sqrt{25-t^{2}}$

f(-1) = $-1\sqrt{25-(-1)^{2}}$

f(-1) = $-\sqrt{24}$

f(-1) = $-2\sqrt{6}$

f($\sqrt{12.5}$) = $\sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}$

f($\sqrt{12.5}$) = 12.5

f(5) = $5\sqrt{25-5^{2}}$

f(5) = 0

Therefore, absolute maximum is f($\sqrt{12.5}$) = 12.5 and absolute minimum is

f(-1) = $-2\sqrt{6}$.