Question

Determine all x in [−π/2, π/2] at which the graph has horizontal tangents

F'(x) = 4cos^3(x)/(sin^2(x)+1)^2


I already found the derivative but im stuck when I make the numerator = 0 and I cant find the horizontal tangents.. Help please

Answer 1

What can be said about the derivative of a function(given) at a point with the gradient of its tangential line?

Answer 2

that is equal to 0?

Answer 3

No. However 0 is the gradient of a horizontal line. In your case, the gradient of the tangent is then 0. But how will that help you find x? You have the derivative of a curve function. Can you recall the relationship between the 2?

Answer 4

Between the derivative of a function and the gradient of the function's tangential line. Hint: what do you do when you derive, in this case, dy/dx?

Answer 5

u make it equal to zero? In order to find the horizontal tangent line... Im just stuck with my algebra when i make it equal to zero

Answer 6

Can you show me what you did?

Answer 7

4cos^3x=0 then im stuck here

Answer 8

Hmm I can't do it myself for now, but here is what I got from somewhere:

(4 cos^3(x))/(1+sin^2(x))^2 = 0
Divide both sides by a constant to simplify the equation.
Divide both sides by 4:
(cos^3(x))/(1+sin^2(x))^2 = 0
Solve for x in the numerator.
Multiply both sides by (1+sin^2(x))^2 assuming (1+sin^2(x))^2 !=0:
cos^3(x) = 0 for 1+sin^2(x)!=0
Eliminate the exponent.
Take the cube root of both sides:
cos(x) = 0 for 1+sin^2(x)!=0
Eliminate the cosine from the left hand side.
Take the inverse cosine of both sides:
x = pi/2+pi n for 1+sin^2(x)!=0 and n element Z
Omit unnecessary assumptions.
The roots x = pi/2+pi n never violate 1+sin^2(x)!=0, which means this assumption can be omitted:
Answer: |
| x = pi/2+pi n for n element Z

Now you will have to figure out how the restriction on x applies to the final answer.

Answer 9

okay thanks