1=2-4e^-(x-1)^2
determine values of x
I have gotten to:
-(x-1)^2=ln(1/4)
and now not sure what to do

Question

1=2-4e^-(x-1)^2
determine values of x
I have gotten to:
-(x-1)^2=ln(1/4)
and now not sure what to do

anonymous · Answer

$$1=2-4e ^{-(x-1)^2}$$

mathstudent55 · Answer

$\large 1=2-4e ^{-(x-1)^2}$

Subtract 2 from both sides. Then divide both sides by -4.

anonymous · Answer

I am at: -(x-1)^2=ln(1/4)

mathstudent55 · Answer

$\large \dfrac{1}{4}= e ^{-(x-1)^2}$

$\large \ln\dfrac{1}{4}= \ln e ^{-(x-1)^2}$

$\large \ln\dfrac{1}{4}= -(x-1)^2$

$\large -(x-1)^2 = \ln\dfrac{1}{4} $

Good. I got there too.
Now multiply both sides by -1.
Then take the square root of both sides being careful to set the right side as +/- the square root. Then add 1 to both sides of both equations.
as in: $x^2 = k$, then $x = \pm \sqrt{k} $

mathstudent55 · Answer

Also, $ln \dfrac{1}{4} = \ln 2^{-2} = -2 \ln 2$.
Which may be a more elegant way of writing the log part.

anonymous · Answer

when I multiply be negative 1 does that give me (x-1)^2=??

mathstudent55 · Answer

Yes.

$\large -(x-1)^2 = -2\ln 2$

$\large (x-1)^2 = 2\ln 2$

anonymous · Answer

ahhhhh ok so it could also be (x-1)^2=ln4

mathstudent55 · Answer

Yes.

mathstudent55 · Answer

Now take the square root of both sides, and use the +/- on the right side before the root sign.
Then add 1 to both sides.

anonymous · Answer

$$x=\pm \sqrt{\ln4}+1$$

mathstudent55 · Answer

Correct.

$\large x-1 = \pm \sqrt{2\ln 2}$

$\large x =1 \pm \sqrt{2\ln 2}$

mathstudent55 · Answer

Your answer is fine.

anonymous · Answer

Now I need to find the derivative of $$y=2-4e ^{-(x-1)^{2}}$$

anonymous · Answer

I have a mental blank

anonymous · Answer

is it $$-4e ^{-(x-1)^{2}}something?$$

mathstudent55 · Answer

u is a function of x
Then:
$\large \dfrac{d}{dx} e^u = e^u u'$

mathstudent55 · Answer

$\dfrac{d}{dx} e^x = e^x \cdot \dfrac{d}{dx} x$, but $\dfrac{d}{dx} x = 1$, so

$\dfrac{d}{dx} e^x = e^x $

mathstudent55 · Answer

In your case, the exponent is not simply x, so you need to take its derivative too.

mathstudent55 · Answer

$\large y=\color{red}{2}-4e ^{\color{green}{-(x-1)^{2}}}$

$\large y'=\color{red}{0}-4e ^{-(x-1)^{2}}\color{green}{(-1)(2)(x - 1)}$

$\large y'=8(x - 1)e ^{-(x-1)^{2}}$

mathstudent55 · Answer

I hope the colors make it clearer.

anonymous · Answer

Thank you so very very much for all of your help, I really appreciate it :)

mathstudent55 · Answer

You're very welcome.