Answer: Absolute minimum: f(-1) = -2[tex]\sqrt{6}[/tex]
Absolute maximum: f([tex]\sqrt{12.5}[/tex]) = 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) = [tex]t\sqrt{25-t^{2}}[/tex]
f'(t) = [tex]1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)[/tex]
f'(t) = [tex]\sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }[/tex]
f'(t) = [tex]\frac{25-2t^{2}}{\sqrt{25-t^{2}} }[/tex] = 0
For this function to be zero, only denominator must be zero:
[tex]25-2t^{2} = 0[/tex]
t = ±[tex]\sqrt{2.5}[/tex]
[tex]\sqrt{25-t^{2}}[/tex] ≠ 0
t = ± 5
Now, evaluate critical points in the given interval.
t = [tex]-\sqrt{2.5}[/tex] and t = - 5 don't exist in the given interval, so their f(x) don't count.
f(t) = [tex]t\sqrt{25-t^{2}}[/tex]
f(-1) = [tex]-1\sqrt{25-(-1)^{2}}[/tex]
f(-1) = [tex]-\sqrt{24}[/tex]
f(-1) = [tex]-2\sqrt{6}[/tex]
f([tex]\sqrt{12.5}[/tex]) = [tex]\sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}[/tex]
f([tex]\sqrt{12.5}[/tex]) = 12.5
f(5) = [tex]5\sqrt{25-5^{2}}[/tex]
f(5) = 0
Therefore, absolute maximum is f([tex]\sqrt{12.5}[/tex]) = 12.5 and absolute minimum is
f(-1) = [tex]-2\sqrt{6}[/tex].
