What are the critical values of #f(x)=x-xsqrt(e^x#?

Question

Cameron Alexander · Accepted Answer

#x=0#

First of all, your function is well defined on #\mathbb{R}# and also its derivative. The critical values are the set of point where #f'(x) = 0#. In this case, you have

#f'(x) = 1- sqrt(e^x) - \frac{1}{2}xsqrt(e^x)#.

So, you have to solve

#1- sqrt(e^x) - \frac{1}{2}xsqrt(e^x)=0 \quad# which is equivalent to #\quad e^(x/2)(x+2)=2#. Now this is a nonlinear equation. But you could think in this way. You want to find #x# such that #\quad e^(x/2)(x+2) - 2=0#. But for which #x# we have

#\quad e^(x/2)(x+2) - 2>0#? An easy calculation gives us

#x > 0 \Rightarrow\quad e^(x/2)(x+2) - 2>0# and #x < 0 \Rightarrow\quad e^(x/2)(x+2) - 2<0#. So the only choice you have is #x=0#. We see that #f'(0) = 1-sqrt(e^0)-1/2 0sqrt(e^0)=0.# So the only critical point is #0#.

Isabella Ackerman · Answer

undefined To find the critical values of the function $ f(x) = x - x\sqrt{e^x} $, we first need to find its derivative, set it equal to zero, and solve for $ x $.

Given $ f(x) = x - x\sqrt{e^x} $,

1. Find the derivative of $ f(x) $:
$$ f'(x) = 1 - \sqrt{e^x} - \frac{x}{2\sqrt{e^x}} $$

2. Set $ f'(x) $ equal to zero and solve for $ x $:
$$ 1 - \sqrt{e^x} - \frac{x}{2\sqrt{e^x}} = 0 $$

To simplify, let $ u = \sqrt{e^x} $:
$$ 1 - u - \frac{\ln(u^2)}{2} = 0 $$

Multiplying both sides by 2 to clear the fraction:
$$ 2 - 2u - \ln(u^2) = 0 $$

Using properties of logarithms:
$$ 2 - 2u - 2\ln(u) = 0 $$

Rearranging terms:
$$ 2\ln(u) - 2u = 2 $$

Dividing by 2:
$$ \ln(u) - u = 1 $$

This equation cannot be solved algebraically. Therefore, we'll need to use numerical or graphical methods to find its solutions.

Once you find the solutions for $ x $, those values will be the critical values of the function $ f(x) $.