Find the critical numbers of fx 6 sin x 6 cos x x fx fx

Find the critical numbers of

f(x) = 6 sin x + 6 cos x

x =

f(x) =

f(x) =

(Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

critical number(s)	x =
absolute maximum value	f(x) =
absolute minimum value	f(x) =

Find the critical numbers of f(x) = 6 sin x + 6 cos x and determine the extreme values on [0,pi /2] (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

Solution

f(x) = 6sinx + 6cosx

Deriving :

f\'(x) = 6cosx - 6sinx

To find criticals, we have to equate derivative to 0

6cosx - 6sinx = 0

6cosx = 6sinx

cosx = sinx

We know this happens only at x = pi/4, considering given interval [0 , pi/2]

So, x = pi/4 ---> critical value ---> FIRST ANSWER

Now, we have to find the value of the function at the critical value and at the endpoints to figure out the max and min

f(0) = 6sin0 + 6cos0 = 0 + 6 = 6

f(pi/4) = 6sin(pi/4) + 6cos(pi/4) ---> 6(sqrt2/2) + 6(sqrt2/2) ---> 6sqrt(2)

f(pi/2) = 6sin(pi/2) + 6cos(pi/2) --> 6 + 0 ---> 6

So, here are the answers :

critical numbers, x = pi/4 ---> ANSWER
Absolute maximum value f(x) = 6*sqrt(2) at x = pi/4 ---> ANSWER
Absolute minimum value f(x) = 6 at x = 0 , pi/2 ---> ANSWER