Use the first derivative test to find any relative extrema

Question

Hello could someone please help me find the extrema on this function:

 

f(x)= (3/2)sin((πx/2)-1) on the interval [0,4]

 

I did the first derivative and I think it's f'(x)= (3/2)cos((πx/2)-1)(π/2)

 

I can't seem to figure out what would make this function a zero. Thanks! 

Answer 1

f'(x) = 3π/4[cos(πx/2 - 1)]

 

That equals zero when cos(πx/2 - 1) = 0

 

So, πx/2 - 1 = arc cos 0. 

 

That means πx/2 = (arc cos 0) + 1

 

And x = 2/π [(arc cos 0) + 1]

 

Now what are the values of arc cos 0 that produce all possible x values in the interval from [0,4]?

 

We can replace arc cos 0 by values like π/2, 3π/2, 5π/2, 7π/2, 9π/2.... but you will stop as soon as we end up outside that interval.

 

x₁ = 2/π [ π/2 + 1] = 1 + 2/π.

x₂ = 2/π [3π/2 + 1] = 3 + 2/π.

 

The next substitution will take us outside the [0,4] interval, so those x-values are the critical numbers.

 

Check out: https://www.desmos.com/calculator/ru8wcf0w1m

 

You might want to determine the value of f(0) and f(4) to see if an absolute max/min happens at the endpoints of the interval. The graph shows that the x₁ produces a relative max and x₂ produces a relative min.

Answer 2

Rebecca,

         So far so good...when is the function cos(θ)=0?, Hint when  θ=n*pi -pi/2 n=0,1,2....

So when (pi*x/2)-1=(n*pi/2)-pi/2 to find the values solve for x. If you find a graph of the cosine you will instantly see where it is zero.

Hope this helps

 

Jim