Find all points on graph of f(x)=2sinx+sin^2x where slope = 0

[illjazz]

Homework Statement

Find all points on the graph of the function

$$f(x)=2sinx+sin^2x$$

at which the tangent line is horizontal.

Homework Equations

- Power rule
- Chain rule
- Product rule?

The Attempt at a Solution

So I want all points at which the tangent line to this function has a slope of zero.

$$f(x)=2sinx+sin^2x$$

$$f'(x)=2cosx+(sinx)^2$$

$$f'(x)=2cosx+2sinx*\frac{d}{dx}sinx$$

$$f'(x)=2cosx+2sinx*cosx$$

$$f'(x)=cosx(2+2sinx)$$

...?

How do I go from here? I know to set that last equation equal to zero and find all the solutions, but just how do I find those solutions?

The solution from the book says:
$$(\frac{\pi}{2}+2n\pi,3),(\frac{3\pi}{2}+2n\pi,-1)$$ where n an integer

 

[rock.freak667]

Slope=0 means f'(x)=0

then you use the general rule that if AB=0 then,A=0,B=0

 

[illjazz]

Slope=0 means f'(x)=0

then you use the general rule that if AB=0 then,A=0,B=0

Right, I know.. but I don't see how to arrive at those values given in the solution. Is my calculator required here?

 

[adartsesirhc]

First, think back to algebra when you had those equations like (x-4)(x-2)=0, and you solved them by setting up two possibilities: either x-4=0, or x-2=0. You can use the same principle here.

If $$cosx=0$$, or if $$2+2sinx=0$$, then $$f'(x)=0$$.

Now simply find all the points where this can happen. You don't need a calculator - simple trigonometry can help. Where is $$cosx=0$$? And where is $$sinx=-1$$?

 

[tiny-tim]

Find all points on the graph of the function

$$f(x)=2sinx+sin^2x$$

at which the tangent line is horizontal.
…
The solution from the book says:
$$(\frac{\pi}{2}+2n\pi,3),(\frac{3\pi}{2}+2n\pi,-1)$$ where n an integer

.. but I don't see how to arrive at those values given in the solution. Is my calculator required here?

Hi illjazz!

(have a pi: π and a squared: ² )

I think you're misunderstanding the notation in the solution.

sin(π/2 + 2nπ) = sin(π/2) = 1, and so f(π/2 + 2nπ) = 3.

sin(3π/2 + 2nπ) = sin(3π/2) = … ?, and so f(3π/2 + 2nπ) = … ?.

 

[HallsofIvy]

Homework Statement

Find all points on the graph of the function

$$f(x)=2sinx+sin^2x$$

at which the tangent line is horizontal.

Homework Equations

- Power rule
- Chain rule
- Product rule?

The Attempt at a Solution

So I want all points at which the tangent line to this function has a slope of zero.

$$f(x)=2sinx+sin^2x$$

$$f'(x)=2cosx+(sinx)^2$$

No, it is not. I know what you mean but this is terrible notation. You mean
$$f'(x)= 2 cosx+ \frac{d}{dx} (sin x)^2$$

$$f'(x)=2cosx+2sinx*\frac{d}{dx}sinx$$

$$f'(x)=2cosx+2sinx*cosx$$

Okay, now that is correct.

$$f'(x)=cosx(2+2sinx)$$

But why did you do this? Just solve 2cos x+ 2 sin x cos x= 0. That is the same as
2 cos x (1+ 2 sin x)= 0 which means either cos x= 0 or 1+ 2 sin x= 0. That is easy to solve.

...?

How do I go from here? I know to set that last equation equal to zero and find all the solutions, but just how do I find those solutions?

The solution from the book says:
$$(\frac{\pi}{2}+2n\pi,3),(\frac{3\pi}{2}+2n\pi,-1)$$ where n an integer