How Do You Find Points with Horizontal Tangents on f(x) = 2sinx + (sinx)^2?

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SUMMARY

The discussion focuses on finding points with horizontal tangents for the function f(x) = 2sin(x) + (sin(x))^2. The key steps involve calculating the derivative f'(x) = 2cos(x)(sin(x) + 1) and setting it to zero to find critical points. The solutions to the equation 2cos(x)(sin(x) + 1) = 0 yield x-values where the tangent is horizontal, specifically at x = π/2, (3π)/2, and their periodic extensions. Corresponding y-values are then determined by substituting these x-values back into the original function.

PREREQUISITES
  • Understanding of derivatives and the concept of tangent lines
  • Familiarity with trigonometric functions and their properties
  • Knowledge of solving equations involving trigonometric identities
  • Ability to substitute values into functions to find corresponding outputs
NEXT STEPS
  • Learn how to apply the Chain Rule in calculus for more complex functions
  • Study the properties of trigonometric functions and their derivatives
  • Explore the concept of critical points and their significance in graphing functions
  • Investigate periodic functions and their implications in calculus
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Students studying calculus, particularly those focusing on derivatives and tangent lines, as well as educators looking for examples of finding horizontal tangents in trigonometric functions.

nejnadusho
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Homework Statement



Find all points on the graph of the function at which the tangent line is horizontal.

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

Homework Equations


Hi guys I have an issue. I do not know how to approach this problem.

I know the chain rule but I still do not know how to solve the prolem.

Thank you for your help.

The Attempt at a Solution

 
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the tangent line is the slope of f(x) at some point, how do you find that slope?
 
The derivative of the function gives you the slope of the tangent line at each point x.

How do you know when a line is horizontal?

Now, how can you find which points have horizontal lines?
 
Last edited:
what is the slope of a horizontal line?
 
slope is always rise over run; if its instantaneous then its derivative of rise of derivative of run.
if its an entire straight line, the slope is rise over run. So, what's the rise per unit of run?
 
nejnadusho, the ball is back in your court! Everyone is asking you the same question: what is the slope of a horizontal line?
 
Ok
I am sorry there was something wrong and I could not open the page yestarday.
I guess I should find the derivative of the f(x) which means I should find f'(x) and after
the equation of the tangent line . right?

And i guess the slope is horizontal when the f(x) of the slope equals ohhh no.
When x of the function of the slpope is 0(zero) right?
 
nejnadusho said:
Ok
I am sorry there was something wrong and I could not open the page yestarday.
I guess I should find the derivative of the f(x) which means I should find f'(x) and after
the equation of the tangent line . right?

And i guess the slope is horizontal when the f(x) of the slope equals ohhh no.
When x of the function of the slpope is 0(zero) right?

I hope that means 'when f'(x)=0'.
 
Actually
I got lost again because I found the derivative
but How am I suppose to find the equation of the tangent line when I do not have a certain point.

Do I need it actually?
I mean some point?

So if I substitute x by 0 I am going to find the point where the slope of the tangent line is 0 ?
 
Last edited:
  • #10
nejnadusho said:
Actually
I got lost again because I found the derivative
but How am I suppose to find the equation of the tangent line when I do not have a certain point.

Do I need it actually?
I mean some point?

It doesn't ask you for the equation of the tangent line. It's asks you for what values of x and y does the tangent line have zero slope. Start by finding the x values.
 
  • #11
Tell us what you got for f'(x). Write the equation f'(x)=0. Now try to solve for x.
 
  • #12
So if I substitute x by 0 I am going to find the point where the slope of the tangent line is 0 ?

But I am going to find only one y for each x so how am I suppose to find more than one zero slope if there is?
 
  • #13
If you 'substitute x by 0' you are going to find the slope of the tangent line at x=0. You want to know where the slope is zero. Tell us what you got for f'(x). Write the equation f'(x)=0. Now try to solve for x.
 
  • #14
ok
f'(x)= 2cos x (sin x + 1)
2cos x (sin x + 1) = 0
 
  • #15
ok but how am I suppose to solve it for x now?
 
  • #16
nejnadusho said:
ok
f'(x)= 2cos x (sin x + 1)
2cos x (sin x + 1) = 0

Good! You've even factored it. If the product of those factors is 0, then either cos(x)=0 OR (sin(x)+1)=0. For what values of x is cos(x)=0?
 
  • #17
Pi/2 for cosx

and ((3*Pi)/2) for sin
 
  • #18
I am dum

I stay I am doing something and I cannot realize that I just solve the problem.
 
  • #19
Good so far, but the problems asks for ALL points. E.g. cos(3pi/2)=0, cos(5pi/2)=0, etc. etc. And sin(-pi/2)=(-1) and sin(7pi/2)=(-1). Can you clearly describe all of them? Once you've done that, you just need to get the corresponding y values.
 
  • #20
Thank you very much for the way you helped me.
Thank you very much guys.
 
  • #21
Yes but isn't the coresponding values of y allways the same?

are not they always y= 0 for cos and y = -1 for sin ?

I mean for this problem.?
 
  • #22
To get the y values you have to put the x values into f(x) = 2sinx + (sinx)^2, not the derivative, right?
 
  • #23
ohhh really?

sooo
for f((3*Pi)/2) = -1
and
f(Pi/2) = 3


but why all
that
why I take the values from the derivative and substitute them in the original equation?
 
  • #24
so are not these values actually the domain of the original function?
 
  • #25
nejnadusho said:
ohhh really?

sooo
for f((3*Pi)/2) = -1
and
f(Pi/2) = 3


but why all
that
why I take the values from the derivative and substitute them in the original equation?

Because you problem asked you to "Find all points on the graph" and points have both x and y values. Solving f '(x)= 0 gives you the x values where the tangent is 0 but you still need the corresponding y values.
 
  • #26
its just like solving for the zero's (x-intercepts) of any polynomial equation.
how can you get 2cosx*(sinx + 1) to = 0. either cosx = 0 or sinx + 1 = 0. what values of x give you those? and here we see the 2 solutions (actually infinite solutions, but for practical purposes, 2)
 

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