What are the inflections points of y = e 2 x - e x ?

Question

What are the inflections points of           y      =              e                  2          x                    -              e        x            ?

Peruvianoe4p · Accepted Answer

By the Chain Rule, the first derivative is

y' = 2 e^{2 x} - e^{x}

and the second derivative is

y'' = 4 e^{2 x} - e^{x}

. Inflection points occur at x values where the sign of the second derivative changes (from positive to negative or negative to positive).
Setting

y'' = 0

leads to

4 e^{2 x} - e^{x} = 0

. The left-hand side of this equation can be factored as

e^{x} (4 e^{x} - 1) = 0

. Since

e^{x}

is never zero, it follows that we just need to solve

4 e^{x} - 1 = 0

to get

x = \ln (\frac{1}{4}) = - \ln (4) \approx - 1.386

.
You can check that

y'' = 4 e^{2 x} - 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 (\frac{1}{4})} - e^{\ln (\frac{1}{4})} = \frac{1}{16} - \frac{1}{4} = - \frac{3}{16}

. The inflection point is therefore

(- \ln (4), - \frac{3}{16})

.

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

Answered question

Answer & Explanation

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