Thread: horizontal tangent line

Will you please find all values of x for the given function where the tangent line is horizontal?

f(x) = x / (x^2 + 1)^3

Two things: how do you find the slope of the tangent lines in general for f(x)? If a line is horizontal, what is its slope? How can you use these two things together to solve?

If that doesn't help I'll get you started.

Thank you for replying me
If slope is tangent, that means f'(x) = 0. And its slope is constant.
Am I right?

Originally Posted by Matho

Thank you for replying me
If slope is tangent, that means f'(x) = 0. And its slope is constant.
Am I right?

Not quite. Well you might be right and I just am interpreting your post wrong, but I'll post my comment anyway. Disregard if you know this already.

The slope of a line that is tangent to a certain point on f(x) is different at different points because f(x) increases and decreases at different rates. Taking f'(a) will tell you the slope of a line that is tangent to the point (a, f(a)) on the graph of f(x), but the slope of the point (a+1) is going to be different.

You are right though that you need to find when f'(x)=0 and solve for x.

Hi Jameson,

I got f'(x) = (x^6 - x^5 + 3x^4 - 2x^3 + 3x^2 - x + 1) / (x^2 + 1)^5
However, I don't know how to get f'(x) = 0 from this...