For what value(s) of x does g(x) have a horizontal tangent?

[JSmith]

For what value(s) of x does g(x) have a horizontal tangent?
So I know to solve this I just calculate the derivative and then solve for what values the derivative is equal to zero. I think I am making a mistake however...

 

[srmichael]

For what value(s) of x does g(x) have a horizontal tangent?
So I know to solve this I just calculate the derivative and then solve for what values the derivative is equal to zero. I think I am making a mistake however...

Your error was in the 4th step. When you distributed you said \(\displaystyle x(-0.5x^{-1.5})=-0.5x^{0.5}\) when in fact it equals \(\displaystyle -0.5x^{-0.5}\)

Personally I would have just done the quotient rule:

\(\displaystyle g(x)=\frac{x+1}{\sqrt{x}}\)

\(\displaystyle g'(x)=\frac{(1)\sqrt{x}-(x+1)\frac{1}{2\sqrt{x}}}{x}\)

\(\displaystyle g'(x)=\frac{2\sqrt{x}\sqrt{x}-(x+1)(\frac{1}{2\sqrt{x}})2\sqrt{x}}{(x)2\sqrt{x}}\)

\(\displaystyle g'(x)=\frac{2x-(x+1)}{2x^{\frac{3}{2}}}\)

\(\displaystyle g'(x)=\frac{x-1}{2x^{\frac{3}{2}}}\)

\(\displaystyle \frac{x-1}{2x^{\frac{3}{2}}}=0\)

\(\displaystyle x-1=0\)

\(\displaystyle x=1\)

 

[stapel]

Thank you for showing your steps so nicely! I've inserted an unstated step in your work shown:

\(\displaystyle \mbox{a) }\, g(x)\, =\, \frac{x\, +\, 1}{\sqrt{x}}\)

\(\displaystyle \mbox{b) }\,g(x)\, =\, (x\, +\, 1)x^{-0.5}\)

\(\displaystyle \mbox{c) }\,g'(x)\, =\, (x\, +\, 1)\left(-0.5x^{-1.5}\right)\, +\, \left(x^{-0.5}\right)(1)\)

\(\displaystyle \mbox{d) }\,g'(x)\, =\, (x^{1.0})\left(-0.5x^{-1.5}\right)\, +\, (1)\left(-0.5x^{-1.5}\right)\, +\, (1)\left(x^{-0.5}\right)\)

\(\displaystyle \mbox{e) }\,g'(x)\, =\, -0.5x^{-0.5}\, -\, 0.5x^{-0.5}\, +\, x^{-0.5}\)

What happened between lines (d) and (e)? How did you get the power on that middle term?

 

[JSmith]

Awesome. Thanks so much for your help.