## Inflections

For the curve represented by the equation below, find the point of inflection(s).

##
__
__**Hint**

**Hint**

First, find the second derivative of
$$f(x)$$
.

##
__
__**Hint 2**

**Hint 2**

Set
$$f''(x)=0$$
to solve for the inflection point.

An inflection point is a point on the curve/graph at which concavity changes, and occurs when
$$f''(x)=0$$
.

Using the power rule for the first derivative and applying it twice, we’ll get the second derivative power rule:

$$$\frac{d^2}{dx^2}[x^n]=\frac{d}{dx}\frac{d}{dx}[x^n]=\frac{d}{dx}[nx^{n-1}]=n\frac{d}{dx}[x^{n-1}]=n(n-1)(x^{n-2})$$$

Thus, the second derivative is:

$$$f''(x)=e^x-2(2)(2-1)(x^{2-2})$$$

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

$$$f''(x)=e^x-(4)(1)(1)=e^x+4$$$

Note that
$$\frac{d(e^u)}{dx}=e^u\frac{du}{dx}$$
, which is why
$$e^x$$
remains unchanged after performing the second derivative. Solving for
$$x$$
when
$$f''(x)=0$$
to get the x-component inflection point:

$$$e^x-4=0$$$

$$$ln(e^x)=ln(4)$$$

$$$x=1.39$$$

The inflection point consists of both a
$$x$$
and
$$y$$
coordinate. We have solved the
$$x$$
component, but still need to determine the
$$y$$
component. To find
$$y$$
, plug the
$$x$$
component back into the original function:

$$$y=e^{1.39}-2(1.39)^2$$$

$$$y=4.01-2(1.93)=4.01-3.86=0.15$$$

Thus, the inflection point is at (1.39, 0.15)

(1.39, 0.15)