What are the inflections points of #y= e^(2x) - e^x #?

Question

Emma Aldridge · Accepted Answer

By the Chain Rule, the first derivative is #y'=2e^{2x}-e^{x}# and the second derivative is #y''=4e^{2x}-e^{x}#.  Inflection points occur at the values of #x# where the second derivative changes sign (from positive to negative or negative to positive). Setting #y''=0# leads to #4e^{2x}-e^{x}=0#.  The left-hand side of this equation can be factored as #e^{x}(4e^{x}-1)=0#.  Since #e^{x}# is never zero, it follows that we just need to solve #4e^{x}-1=0# to get #x=ln(1/4)=-ln(4)\approx -1.386#. You can check that #y''=4e^{2x}-e^{x}# changes sign at this point by graphing it.  Therefore, there is an inflection point at #x=-ln(4)#.  The second coordinate of this point can be found by plugging it into the original function to get #e^{2\cdot ln(1/4)}-e^{ln(1/4)}=1/16-1/4=-3/16#.  The inflection point is therefore #(-ln(4),-3/16)#.

William Abdalla · Answer

undefined To find the inflection points of $ y = e^{2x} - e^x $, we first find the second derivative and set it equal to zero to find the points of inflection.

First derivative:
$$ \frac{dy}{dx} = 2e^{2x} - e^x $$

Second derivative:
$$ \frac{d^2y}{dx^2} = 4e^{2x} - e^x $$

Setting the second derivative equal to zero:
$$ 4e^{2x} - e^x = 0 $$

Dividing both sides by $ e^x $:
$$ 4e^x - 1 = 0 $$

Adding 1 to both sides:
$$ 4e^x = 1 $$

Dividing both sides by 4:
$$ e^x = \frac{1}{4} $$

Taking the natural logarithm of both sides:
$$ x = \ln\left(\frac{1}{4}\right) $$

Solving for $ x $:
$$ x = -\ln(4) $$

Thus, the inflection point is $ x = -\ln(4) $.