2

The curve with equation $y = \frac {e^{2x}} {4 + e^{3x}}$ has one stationary point. Find the exact values of the coordinates of this point.

I got to the point where this is my $\frac {dy} {dx}$:$$\frac{ (4 + e^{3x}) (2e{^{2x}})-e^{2x}(3e^{3x})}{(4+e^{3x})^2} = 0$$

Is this correct to find the stationary points, if it is, how do I get $x $ from this equation?

imranfat
  • 10,029
Riduan Gonzalez
  • 179
  • Good exercise. So do you know the Quotient Rule? Do you know the Chain Rule regarding e-powers? Show some work, we can help – imranfat Apr 08 '17 at 18:02
  • There, I put where I am stuck – Riduan Gonzalez Apr 08 '17 at 18:03
  • Good, I edited the square on the denominator, but that makes no difference now. You need to set the numerator equal to zero for stationary points. This becomes an e-power equation – imranfat Apr 08 '17 at 18:05
  • I presume that you know that $\frac{p}{q}= 0$ if and only if p= 0. Here that means that $3e^{5x}- 8e^{2x}- 2e^{5x}= e^{5x}- 8e^{2x}= 0$. That is the same as $e^{5x}= 8e^{2x}$ or $e^{3x}= 8$. – user247327 Apr 08 '17 at 18:07
  • Don't forget to find the value $y$ once you have $x$ to get the complete coordinate. – Shaun Apr 08 '17 at 18:11
  • 1
    He "flipped" the numerator part...edited – imranfat Apr 08 '17 at 18:17

5 Answers5

4

Developing your numerator, that must be zero, you have the following: $$ 2e^{5x} + 8e^{2x} - 3e^{5x} = 0 \iff e^{5x} = 8e^{2x} \iff 8 = \frac{e^{5x}}{e^{2x}} = e^{3x} \iff \ln 8 = \ln e^{3x} = 3x\ln e = 3x \iff x = \frac{\ln 8}{3} = \frac{\ln2^3}{3} = \frac{3\ln 2}{3} = \ln 2 $$ Notice that you can divide by an exponential because $e^x\neq 0\;\;\forall x\in\mathbb{R}$.

Abelois
  • 965
2

$\frac {dy} {dx} = \frac{e^{2x}(3e^{3x}) - (4 + e^{3x}) (2e{^{2x}})}{(4+e^{3x})^2}$

Now,

$$\frac {dy} {dx} = 0$$

$$\frac{e^{2x}(3e^{3x}) - (4 + e^{3x}) (2e{^{2x}})}{(4+e^{3x})^2} = 0$$

$$3e^{5x} - 8e^{2x} - 2e^{5x} = 0$$

$$e^{5x} - 8e^{2x} = 0$$

$$e^{2x}(e^{3x} - 8) = 0$$

Either $$e^{2x} = 0$$

Or $$e^{3x} - 8 = 0$$

Now find $x$ and then put in equation to find $y$.

Kanwaljit Singh
  • 8,610
1

Note the quotient rule:$$\frac{d}{dx} \bigg(\frac{f(x)}{g(x)}\bigg) = \frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{g(x)^2}$$

Then we have $$h(x) = \frac{e^{2x}}{4+e^{3x}},\quad f(x)=e^{2x},\quad g(x) = 4+e^{3x}$$

Then $$h'(x) = \frac{2e^{2x}(4+e^{3x})-e^{2x}\cdot 3e^{3x}}{(4+e^{3x})^2}$$

We set $h'(x) = 0$ to find the stationairy points:

$$ \frac{2e^{2x}(4+e^{3x})-e^{2x}\cdot 3e^{3x}}{(4+e^{3x})^2}=0\implies 8e^{2x}+2e^{5x}-3e^{5x}=0$$

Then $$e^{5x}-8e^{2x}=0\implies e^{2x}(e^{3x}-8)=0 \implies e^{3x}=8$$ Note this is the only solution since $e^{2x}\neq 0$. Then:

$$x=\ln 8\implies x=\frac{1}{3}\ln8 = \ln 2$$ which is the only stationary point.

mrnovice
  • 5,773
1

First we want to know the derivative of $e^{2x}$ using the chain rule and $u = 2x$ and $y= e^u$ $$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$ $$y' = e^u(2) = 2e^{2x}$$

Applying the quotient rule $(\frac{u}{v})' = \frac{u'v - v'u}{v^2}$ $$y' = \frac{2e^{2x}(4+ e^{3x}) - 3e^{3x}(e^{2x})}{(4+ e^{3x})} = 0$$

This is only zero when the numerator is zero so $$2e^{2x}(4+ e^{3x}) - 3e^{3x}(e^{2x}) = 0$$

factoring out the $e^{2x}$ we get $$8 + 2e^{3x} - 3e^{3x} = 0$$ $$e^{3x} = 8 \implies x = \frac{\ln 8}{3} = \ln 2$$

Ziad Fakhoury
  • 2,593
0

I will use this a chance to publicize a trick I like for solving exponential equations quickly:

Assuming you have already set the numerator to zero $$ 2e^{5x}+8e^{2x}-3e^{5x}=0 $$ set $u=e^x$ and find $$ 2u^{5}+8u^{2}-3u^{5}=0\implies u=0,u^3=8\implies u=0,2 $$ plugging $u=e^x$ back in, we have $$ e^x=0\;\text{or}\:e^x=2 $$ since $e^x>0$ always, $x=\ln 2$.

operatorerror
  • 29,103