Find the maximum and minimum values of the function g(θ)= 7(θ)- 9sin(θ) on the interval (0,pi)

Question

I have been working on this question for about an hour, but keep getting it wrong.. I don't know what I am doing wrong. Please help!

Answer 1

g(θ) = 7θ – 9sin(θ)

A local maximum or minimum will occur when the first derivative of g(θ) is equal to zero:

dg/dθ = g’(θ) = 7 – 9cos(θ) = 0

Adding 9cos(θ) to both sides gives

9cos(θ) = 7

Then dividing both sides by 9 gives

cos(θ) = 7/9 which leads to θ = cos⁻¹(7/9)

This is only the value of x where there can be a local minimum or maximum.

We can see if we have a maximum or minimum by evaluating the second derivative of g(θ)

when θ = cos⁻¹(7/9).

If the second derivative is positive, we have a minimum; if it is negative, we have a maximum.

d²g/dθ² = g’’(θ) = 9sin(θ) > 0 for all values of θ on the interval (0, π)

So we have a minimum at θ = cos⁻¹(7/9)

The minimum value of g(θ) can be calculated to be g[cos⁻¹(7/9)] ≈ 0.899

The maximum value of g(θ) occurs at one of the endpoints of the interval,

so we calculate g(θ) at both ends and see which value is greater:

g(0) = 7×0 – 9sin(0) = 0 – 0 = 0

g(π) = 7π – 9sin(π) = 7π

So the maximum value of g(θ) is g(π) = 7π

Answer 2

See the video.