Question

Find the critical numbers of the function. (Enter your answers as a comma-separated list. Use n to denote any arbitrary integer values. If an answer does not exist, enter DNE.) f(θ) = 10 cos(θ) + 5 sin2(θ)

Answer 1

If

$f(\theta)=10\cos\theta+5\sin^2\theta$

then the derivative is

$f'(\theta)=-10\sin\theta+10\sin\theta\cos\theta$

Critical points occur where $f'(\theta)=0$. This happens for

$-10\sin\theta+10\sin\theta\cos\theta=0$

$-10\sin\theta(1-\cos\theta)=0$

$\implies-10\sin\theta=0\text{ or }1-\cos\theta=0$

In the first case, we find

$-10\sin\theta=0\implies\sin\theta=0\implies\theta=n\pi$

In the second,

$1-\cos\theta=0\implies\cos\theta=1\implies\theta=2n\pi$

So all the critical points occur at multiples of $\pi$, or $n\pi$. (This includes all the even multiples of $\pi$.)

Answer 2

The critical numbers of the function will be at ($\pi, n\pi$)

Given the function

f(θ) = 10 cos(θ) + 5 sin2(θ)

At the turning point, $\frac{df(\theta)}{d\theta} = 0$

$\frac{df(\theta)}{d\theta} = -10sin \theta + 5(2sin\theta cos\theta)\\\frac{df(\theta)}{d\theta} = -10sin \theta + 10sin\theta cos\theta$

At the turning point,

$-10sin \theta + 10sin \theta cos \theta=0\\-10sin \theta(1-cos\theta) =0\\-10sin\theta = 0 \ and \ 1-cos\theta =0\\sin\theta =0\\\theta=0 + n\pi\\\\For \ 1-cos\theta =0\\cos\theta = 1\\\theta = cos^{-1}1\\\theta = 0 + n\pi$

Hence critical numbers of the function will be at ($\pi, n\pi$)

Learn more here: https://brainly.com/question/14378712