Question

Find the absolute maximum and absolute minimum values of f on the given interval f(t)=t sqrt (4-t^2)

Answer 1

what given interval?

Answer 2

oh sorry [-1,2]

Answer 3

you need to check
\[f(-1),f(2)\] and also the function evaluated at the critical points

Answer 4

\[f'(t)=\sqrt{4-t^2}-\frac{t^2}{\sqrt{4-t^2}}\] i think
add up and get
\[f'(x)=\frac{4-2t^2}{\sqrt{4-t^2}}\]

Answer 5

to find the critical points, set the numerator equal to zero an solve, pretty much in your head you get \(t=\sqrt2\)

Answer 6

How did you combine the derivative to get \[4-t^2/ \sqrt{4-t^2}\]

Answer 7

algebra i guess
common denominator (only denominator) is \(\sqrt{4-t^2}\)

Answer 8

\[\sqrt{4-t^2}-\frac{t^2}{\sqrt{4-t^2}}=\frac{{4-t^2}}{\sqrt{4-t^2}}-\frac{t^2}{\sqrt{4-t^2}}\]

Answer 9

numerator is \(4-2t^2\) not \(4-t^2\)

Answer 10

so when you multiply \[\sqrt{4-t^2}\] to the numerator,you get \[4-2t^2\]?

Answer 11

how do you cancel out the radical from the numerator?