x-coordinates of tangent line

**paguy30** · May 4th 2010, 12:22 PM

I'm a little put off by the wording on this question, if anyone can offer some guidance on how to begin that would be great. also, sorry if i enter the symbols wrong, i'm new here. i don't know why the square root symbol isn't covering all of $x^2 +7$ , but it should.

"find the x-coordinates of the points on the graph where the tangent line is horizontal"
$<br /> f(x) = \frac{x}{\sqrt(x^2 +7)}$

**surjective** · May 4th 2010, 12:26 PM

Hello,

A horizontal tangentline means that:

$f'(x)=0$

All you have to do is:

1) Find the derivative $f'(x)$

2) Solve the equation $f'(x)=0$ with respect to x

3) Done

**paguy30** · May 4th 2010, 12:44 PM

ok so when i'm doing the derivative i use chain rule to figure out the bottom part right?
$<br /> \frac {x}{\sqrt{x^2+7}}$

derivative of top = 1
derivative of bottom using chain rule =

u= $x^2+7$
h(u) = $\sqrt{u}$

$\frac{d}{dx} \sqrt{x^2+7} = 2x*\sqrt{x^2+7} ??$

then use quotient rule for the rest?

$\frac {f\prime(x) * g(x) - f(x) * g\prime(x)}{g(x)^2}$

$\frac{(1)(\sqrt{x^2+7}) - (x)(2x * \sqrt{x^2+7})}{\sqrt{x^2+7}^2}$ ???

then would i just plug in 0 for x???

thanks for your help!

**surjective** · May 4th 2010, 01:06 PM

Hello,

The derivative becomes:

$f'(x)=\frac{\sqrt{x^{2}+7}-x\cdot2x\cdot \frac{1}{2\sqrt{x^{2}+7}}}{(\sqrt{x^{2}+7})^{2}}=\ frac{\sqrt{x^{2}+7}- \frac{x^{2}}{\sqrt{x^{2}+7}}}{(\sqrt{x^{2}+7})^{2} }=\frac{\sqrt{x^{2}+7}}{(\sqrt{x^{2}+7})^{2}} + \frac{\frac{x^{2}}{\sqrt{x^{2}+7}}}{(\sqrt{x^{2}+7 })^{2}}$

This could be simplified to:

$f'(x)=\frac{1}{\sqrt{x^{2}+7}} + \frac{x^{2}}{(\sqrt{x^{2}+7})^{\frac{3}{2}}}$

Now since the tangentline must be horizontal then:

$f'(x)=0$

Which means:

$\frac{1}{\sqrt{x^{2}+7}} + \frac{x^{2}}{(\sqrt{x^{2}+7})^{\frac{3}{2}}}=0$

In this equation you have to find x.

Hope that helpes.

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