What are the points of inflection of f ( x ) = e 2 x - e x ?

Question

What are the points of inflection of             f              (        x        )              =          e              2        x              -          e      x      ?

Devyn Rosario · Accepted Answer

Points of inflection occur when the second derivative is equal to zero. Find this by first differentiating f(x) to get f'(x), then differentiating f'(x) to get f''(x).

f (x) = e^{2 x} - e^{x}

f' (x) = 2 e^{2 x} - e^{x}

f'' (x) = 4 e^{2 x} - e^{x}

Set f''(x) equal to zero to find possible points of inflection:

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

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

Rewrite as a natural log:

x = \ln (4 e^{2 x})

x = \ln 4 + \ln (e^{2 x})

x = \ln 4 + 2 x \ln e

x = \ln 4 + 2 x

- x = \ln 4

x = - \ln 4

x \approx - 1.3863

Check if this is a point of inflection by making sure f''(x) is positive on one side of the x value, and negative on the other (make a sign chart):

- x = - \ln 4 +

Hence,

x = - \ln 4

is the point of inflection of

f (x) = e^{2 x} - e^{x}

What are the points of inflection of f(x)=e^(2x)-e^x?

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