Solving derivative ≥0

_rob

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Nov 28, 2018
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Hi, I need some pointers. I need to solve Df(x)≥0 for the following function.

f(x)=xe{-x^2}+x+1

I have read up on solving derivatives, but still don't understand how to handle the case above.

as I suspect formatting problems with the function, here is the latex version:

f(x)=xe^{-x^2+x+1}
 
What do you get for \(\displaystyle f'(x)\)?
 
I'm able to plot f(x)=xe^{-x^2+x+1} and get a table of values out of it.

For instance
if x=0, f(x)=0.
if x=2, f(x) = 2.71828

But then I don't really know how to continue with the meat of the question.

The actual question translates to something like this:
[FONT=Liberation Serif, serif]Weconsider the function ƒdefined on for [/FONT]f(x)=xe^{-x^2+x+1}[FONT=&quot].Study and solve [/FONT]Df(x)≥0
 
I'm able to plot f(x)=xe^{-x^2+x+1} and get a table of values out of it.

For instance
if x=0, f(x)=0.
if x=2, f(x) = 2.71828

But then I don't really know how to continue with the meat of the question.

The actual question translates to something like this:
Weconsider the function ƒdefined on for f(x)=xe^{-x^2+x+1}.Study and solve Df(x)≥0
Can you calculate the expression for f'(x) - using analytical methods?
 
Hi, I need some pointers. I need to solve Df(x)≥0 for the following function.

f(x)=xe{-x^2}+x+1

I have read up on solving derivatives, but still don't understand how to handle the case above.

as I suspect formatting problems with the function, here is the latex version:

f(x)=xe^{-x^2+x+1}

To follow up, we are given:

\(\displaystyle f(x)=xe^{-x^2+x+1}\)

And so we find:

\(\displaystyle f'(x)=(1)e^{-x^2+x+1}+xe^{-x^2+x+1}(-2x+1)=e^{-x^2+x+1}(1+x(-2x+1))=(-2x^2+x+1)e^{-x^2+x+1}\)

Now, given that \(\displaystyle e^{-x^2+x+1}>0\) for all real \(\displaystyle x\), we observe that:

\(\displaystyle f'(x)\ge0\)

is equivalent to:

\(\displaystyle -2x^2+x+1\ge0\)

or:

\(\displaystyle 2x^2-x-1\le0\)

Factor:

\(\displaystyle (2x+1)(x-1)\le0\)

Now we can see from the unfactored form that the parabolic expression must be non-positive between and including the roots, hence the solution is in the interval:

\(\displaystyle \left[-\dfrac{1}{2},1\right]\)

Let's look at a plot of the given function with the above interval shaded:

fmh_0006.jpg
 
I'm able to plot f(x)=xe^{-x^2+x+1} and get a table of values out of it.

For instance
if x=0, f(x)=0.
if x=2, f(x) = 2.71828
This is not correct

But then I don't really know how to continue with the meat of the question.

The actual question translates to something like this:
Weconsider the function ƒdefined on for f(x)=xe^{-x^2+x+1}.Study and solve Df(x)≥0
f(2)\(\displaystyle \neq\)2.71828
To discuss f '(x) <0 you really should compute (ie have) f '(x)
 
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