find the absolute and local maximum and minimum values off.

$\displaystyle f(t) = 5cos(t) $

-pi/2 ≤ t ≤ 3pi/2

i have no clue how to solve this because cos(-π/2) and cos(3π/2) equal 0 but the answer isn't zero.

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- Nov 18th 2009, 02:13 PMbreakabsolute and local maximum and minimum
find the absolute and local maximum and minimum values of

*f*.

$\displaystyle f(t) = 5cos(t) $

-pi/2 ≤ t ≤ 3pi/2

i have no clue how to solve this because cos(-*π*/2) and cos(3*π*/2) equal 0 but the answer isn't zero. - Nov 18th 2009, 02:40 PMskeeter
absolute extrema don't always occur at endpoints.

based on your prior knowledge of trig alone (w/o calculus), you should be able to sketch the relatively simple graph of $\displaystyle f(t)$.

based on that sketch, the absolute max for $\displaystyle f(t)$ is $\displaystyle f(0) = 5$, and the absolute min is $\displaystyle f(\pi) = -5$