# Thread: Trouble Finding Tangent Lines

1. ## Trouble Finding Tangent Lines

The graph of $y = 6(3x^2)^x$ has two horizontal tangent lines. Find the equations for both of them.
Alright, so my approach will be to find the derivative of the function using logarithmic differentiation or $b^x = e^\ln b^x$. Next, I'll set this derivative equal to zero and solve to obtain the tangent lines, since a tangent line occurs when $f'(x) = 0$. Unfortunately, both attempts have been unsuccessful.

Logarithmic Differentiation

$y = 6(3x^2)^x$

$\ln y = \ln [6(3x^2)^x]$

$\frac{dy}{dx} \cdot \frac{1}{y} = \ln 6 + \ln (3x^2)^x$

$\frac{dy}{dx} \cdot \frac{1}{y} = 0 + x\ln 3x^2$

$\frac{dy}{dx} \cdot \frac{1}{y} = (1 \cdot \ln 3x^2) + x(\frac{1}{3x^2} \cdot 6x)$

$\frac{dy}{dx} \cdot \frac{1}{y} = \ln 3x^2 + \frac{6x^2}{3x^2}$

$\frac{dy}{dx} \cdot \frac{1}{y} = \ln 3x^2 + 2$

$\frac{dy}{dx} = 6(3x^2)^x[\ln 3x^2 + 2]$

Solve $y'$ For 0 To Find Horizontal Tangents Of Original Function

$\ln(3x^2) + 2 = 0$

$\ln(3x^2) = -2$

$e^{\ln(3x^2)} = e^{-2}$

$3x^2 = e^{-2}$

$x^2 = \frac{e^{-2}}{3}$

$\sqrt{x^2} = \frac{\sqrt{e^{-2}}}{\sqrt{3}}$

$x = \pm \frac{1}{e\sqrt{3}}$

Evaluate Original Function At $x$

$y = \ln(3x^2)^x$

$f(\frac{1}{e\sqrt{3}}) = 6(3(\frac{1}{e\sqrt{3}})^2)^{\frac{1}{e\sqrt{3}}}$

$f(\frac{1}{e\sqrt{3}}) = 6(3(\frac{1}{3e^2}))^{\frac{1}{e\sqrt{3}}}$

$f(\frac{1}{e\sqrt{3}}) = 6(\frac{3}{3e^2})^{\frac{1}{e\sqrt{3}}}$

$f(\frac{1}{e\sqrt{3}}) = 6(\frac{1}{e^2})^{\frac{1}{e\sqrt{3}}}$

$f(\frac{1}{e\sqrt{3}}) = 6(e^{-2})^{\frac{1}{e\sqrt{3}}}$

$f(\frac{1}{e\sqrt{3}}) = 6(e^{\frac{-2}{1} \cdot \frac{\sqrt{3}}{3e}})$

$y = 6e^{\frac{-2\sqrt3}{3e}}$

(Substitute $-\frac{1}{e\sqrt{3}}$ to get the other solution)

The answer is said to be:

$y = 6e^{\frac{-2\sqrt3}{3e}}$ and $y = 6e^{\frac{2\sqrt3}{3e}}$

Which confirms that the solution is correct.

Thanks a lot for the help!

2. ## Re: Trouble Finding Tangent Lines

You have calculated the first derivative $f'(x)$ correctly. Indeed, if you have to find the horizontal tangent lines (their slope =0) to the curve therefore you have to solve:
$6(3x^2)^x\cdot \ln[3x^2+2]=0 \Leftrightarrow 6(3x^2)^x=0 \ \mbox{or} \ \ln(3x^2)+2=0$
There're no solutions for $6(3x^2)^x=0$, so you have to solve $\ln(3x^2)+2=0$, can you do that?

Afterwards you have found a point on the curve where you get a horizontal tangent line, so you only have to find the $y-$ coordinate afterwards by substituting the $x-$ value in the equation of the curve.

3. ## Re: Trouble Finding Tangent Lines

It appears that my logarithmic and exponential function manipulation skills are egregious. I need to learn to get comfortable with these. xD

Here's what I tried:

$\ln(3x^2) + 2 = 0$

$\ln(3x^2) = -2$

$e^{\ln(3x^2)} = e^{-2}$

$3x^2 = e^{-2}$

$x^2 = \frac{e^{-2}}{3}$

$\sqrt{x^2} = \frac{\sqrt{e^{-2}}}{\sqrt{3}}$

$x = \pm \frac{1}{e\sqrt{3}}$

I'm relatively sure that I made an error up this point.

$y = \ln(3x^2)^x$

$f(\frac{1}{e\sqrt{3}}) = 6(3(\frac{1}{e\sqrt{3}})^2)^{\frac{1}{e\sqrt{3}}}$

$f(\frac{1}{e\sqrt{3}}) = 6(3(\frac{1}{3e^2}))^{\frac{1}{e\sqrt{3}}}$

$f(\frac{1}{e\sqrt{3}}) = 6(\frac{3}{3e^2})^{\frac{1}{e\sqrt{3}}}$

4. ## Re: Trouble Finding Tangent Lines

$\ln(3x^2) + 2 = 0$

$\ln(3x^2) = -2$

$3x^2 = \frac{1}{e^2}$

$x = \pm \frac{1}{\sqrt{3} \cdot e}$

note that $\sqrt{x^2} = |x|$

5. ## Re: Trouble Finding Tangent Lines

The part that's confusing me is how they're getting

$y = 6e^{\frac{-2\sqrt3}{3e}}$ and $y = 6e^{\frac{2\sqrt3}{3e}}$

as an answer. I've been working on this one for a while now and it's the only one I can't get.

Is that answer a byproduct of using the $b^x = e^\ln b^x$ method before substituting the x values into the original function to find the y values?

Normally, I wouldn't be so aggressive in my inquiries, but multiple people are having the same problem with this one in my class. :x

6. ## Re: Trouble Finding Tangent Lines

Originally Posted by Algebrah

The part that's confusing me is how they're getting

$y = 2e^{\frac{-\sqrt6}{3e}}$ and $y = 2e^{\frac{\sqrt6}{3e}}$

those solutions do not agree with the graph of the original function $y = 6(3x^2)^x$

either the original function or the solutions for the horizontal tangent lines are incorrect.

7. ## Re: Trouble Finding Tangent Lines

Here's the problem:

EDIT: Alright, so it turns out that I copied down the answer to a similar problem.

I finished it. Always a good idea to make sure you're copying down the right thing. Haha.

Thanks everyone!