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(3π/2) equal 0 but the answer isn't zero.
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(3π/2) equal 0 but the answer isn't zero.
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 $\displaystyle f(\pi) = -5$