Find the absolute extrema of the function on the closed interval. y = 4 cos x, [0, 27] Step 1 The absolute extrema of a function on a closed interval will occur at any critical numbers in the interval or at the endpoints of that interval. Begin by differentiating the function with respect to x to identify any critical numbers. y = 4 cos x -4 sin (x) y' = -4 sin(r) To find the critical numbers of y, we find all x-values for which y' = and all x-values for which y' does not exist does not exist Step 2 The derivative exists for all values of x. We therefore find all x-values for which y = 0. Since y = -4 sin x, we need to solve -4 sin x =0 Step 3 In the interval (0, 27), -4 sin x = 0 at x = TI Thus, the critical number is x = Step 4 Evaluate y = 4 cos x at the critical number and at the interval endpoints, 0 and 2n. Left Critical Number Right Endpoint Endpoint y(0) = 4 Y() = y(27) =

Calculus: Early Transcendentals
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Chapter1: Functions And Models
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Please solve for step 5, all steps to problem are listed below, solve all parts for step 5, disregard answers already there, they ARE NOT RIGHT, plewse have correct answers, thank you!!
Find the absolute extrema of the function on the closed interval.
y = 4 cos x, [0, 27]
Step 1
The absolute extrema of a function on a closed interval will occur at any critical numbers in the interval or at the endpoints of that interval.
Begin by differentiating the function with respect to x to identify any critical numbers.
y = 4 cos x
-4 sin (x)
y' =
-4 sin(r)
To find the critical numbers of y, we find all x-values for which y' =
and all x-values for which y' does not exist
does not exist
Step 2
The derivative exists for all values of x. We therefore find all x-values for which y = 0. Since y = -4 sin x, we need to solve
-4 sin x =lo
Step 3
In the interval (0, 27), -4 sin x = 0 at x = I
Thus, the critical number is x = T
Step 4
Evaluate y = 4 cos x at the critical number and at the interval endpoints, 0 and 2r.
Left
Critical
Number
Right
Endpoint
Endpoint
y(0) =
4 y(7) =
y(27) = 4
Transcribed Image Text:Find the absolute extrema of the function on the closed interval. y = 4 cos x, [0, 27] Step 1 The absolute extrema of a function on a closed interval will occur at any critical numbers in the interval or at the endpoints of that interval. Begin by differentiating the function with respect to x to identify any critical numbers. y = 4 cos x -4 sin (x) y' = -4 sin(r) To find the critical numbers of y, we find all x-values for which y' = and all x-values for which y' does not exist does not exist Step 2 The derivative exists for all values of x. We therefore find all x-values for which y = 0. Since y = -4 sin x, we need to solve -4 sin x =lo Step 3 In the interval (0, 27), -4 sin x = 0 at x = I Thus, the critical number is x = T Step 4 Evaluate y = 4 cos x at the critical number and at the interval endpoints, 0 and 2r. Left Critical Number Right Endpoint Endpoint y(0) = 4 y(7) = y(27) = 4
We therefore condude that the absolute maximum is y(0) = y(2z) =
and the absolute minimum is y(a) =
We write the answers as ordered pairs.
minimum
(x, y) = |
4
(x, y) =
-4
(smaller x-value)
maximum
(x, y) =
(larger x-value)
Transcribed Image Text:We therefore condude that the absolute maximum is y(0) = y(2z) = and the absolute minimum is y(a) = We write the answers as ordered pairs. minimum (x, y) = | 4 (x, y) = -4 (smaller x-value) maximum (x, y) = (larger x-value)
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