# What are the points of inflection, if any, of f(x)=12x^3 -16x^2 +x +7 ?

Nov 14, 2017

Find where the second derivative is equal to 0.

#### Explanation:

Points of inflection in a function are those points where the function changes its concavity, between convex (i.e. concave up), and concave (i.e. concave down). Because a positive second derivative denotes a point where the function is convex and a negative one a point where the function is concave, at a point where concavity changes the second derivative must be 0.

Thus, step 1 is finding the second derivative using the Power Rule:

$f \left(x\right) = 12 {x}^{3} - 16 {x}^{2} + x + 7 \to \frac{\mathrm{df}}{\mathrm{dx}} = 36 {x}^{2} - 32 x + 1 \to \frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = 72 x - 32$

Now, we set this equal to 0 to find the appropriate x value or values. Since our second derivative is linear, we know there will be only one appropriate x value, and we know that the second derivative will change from positive to negative or negative to positive at that point. Thus, we know it will be an inflection point.

$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = 72 x - 32 = 0 \to 72 x = 32 \to x = \frac{32}{72} = \frac{16}{36} = \frac{8}{18} = \frac{4}{9}$

The only inflection point is at $x = \frac{4}{9}$, and the y coordinate is...

f(4/9) = 12(4/9)^3 -16(4/9)^2 + 4/9 + 7 = 12(64/729)-16(16/81) + 4/9 + 7 ~5.337

The only point of inflection is at $\left(\frac{4}{9} , 5.337\right)$