## Inflection Point

For the curve represented by the equation below, what value of x does the only point of inflection occurs at?

##
__
__**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:

$$$f''(x)=3(3-1)x^{3-2}+2(2-1)x^{2-2}-0$$$

$$$=3(2)x^{1}+2(1)x^{0}$$$

$$$f''(x)=6x+2(1)=6x+2$$$

Solving for
$$x$$
when
$$f''(x)=0$$
to get the x-component inflection point:

$$$6x+2=0$$$

$$$x=-\frac{2}{6}=-\frac{1}{3}$$$

Since
$$f''(x)=0$$
and
$$f''(x)$$
changes signs at
$$x=-1/3$$
, the inflection point is at
$$x=-1/3$$
.

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