Find the absolute extrema of the function on the closed interval.

Question

Answer 1

abs max (0,1)

abs min (1/6, sqr1.5)

Answer 2

Hello Bennie!

For this question, I would suggest taking the first derivative for f(x) then setting it to 0 to find your critical points and choosing a point between the given interval [0,1/6] to plug into f'(x) to help dictate your absolute extremas.

f'(x) = d/dx(cos(πx)) = -πsin(πx)

Let f'(x) = 0

Then solving for our critical point(s)...

0 = - πsin(πx)

0 = sin(πx)

arcsin(0) = πx

0 = πx

0 = x since we are restricted to the given interval of [0,1/6] from the premise of the question.

Since, x = 0 is one of our endpoints, we just have to pick a single point between [0,1/6] to dictate what kind of absolute extrema it is.

Let's pick x = 1/6 since it can probably tell us if there is an absolute extrema at the other endpoint.

Plugging in x = 1/6 into f'(x) we get...

f'(1/6) = - πsin(π/6)

= - π*(1/2) referencing our unit circle at sin(π/6)

and we can see that the result comes out negative, which means it's decreasing from one endpoint to the other.

If you are not convinced, you can choose another point x between [0,1/6] and f'(x) should still be negative. But, choosing x = 1/6 for this closed interval just happened to be the most negative within that interval which is an indicator that x = 1/6 is an absolute minimum extrema.

With the restriction of the given interval, we can dictate that a location of an absolute max extrema is at:

x = 0

and x = 1/6 is the location of the absolute minimum extrema.

Plugging each back into f(x) will give the y-coordinate will yield the coordinates for the absolute extremas.

f(0) = 1

f(1/6) = (√3)/2

Thus,

The absolute max is: (0,1)

The absolute min is: (1/6, (√3)/2)

Here is the graph for it to help convince you further.

https://www.desmos.com/calculator/zeopm2rtoa

Hope this helps!

Answer 3

f(x) = cos(πx)

The absolute extrema on [0, 1/6] occur at a critical point or at an endpoint.

f'(x) = -πsin(πx) = 0 πx = 0, π, 2π, ... So, x = 0, 1, 2, ... There are no critical points in (0, 1/6), so we need only check the endpoints.

f(0) = cos0 = 1

f(1/6) = cos(π/6) = √3/2 < 1

Absolute max: (0, f(0)) = (0, 1)

Absolute min: (1/6, f(1/6)) = (1/6, √3/2)