Pre-Calculus - Determine Points of Inflection

Determine the values for the points of inflection of the following function:

$\frac{1}{3}$

$\frac{-1}{3}$

Explanation

To solve, you must set the second derivative equal to 0 and solve for x. To differentiate twice, use the power rule as outlined below:

Power Rule: $(x^n)'=nx^{n-1}$

Therefore:

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

Remember, the derivative a constant is 0.

$f''(x)=3*2x^{2-1}-2x^{1-1}=6x-2$

Now, set it equal to 0. Thus,

$x=\frac{2}{6}$

$x=\frac{1}{3}$

Determine the x-coordinate of the inflection point of the function .

$\frac{4}{3}$

$-\frac{4}{3}$

$\frac{8}{3}$

Explanation

The point of inflection exists where the second derivative is zero.

, and we set this equal to zero.

$x=\frac{8}{6}=\frac{4}{3}$

Find the point of inflection of the function .

$\left(\frac{1}{2},\frac{7}{2}\right)$

Explanation

To find the x-coordinate of the point of inflection, we set the second derivative of the function equal to zero.

$x=\frac{6}{12}=\frac{1}{2}$ .

To find the y-coordinate of the point, we plug the x-coordinate back into the original function.

$y=2\left(\frac{1}{2}\right)^3-3\left(\frac{1}{2}\right)^2+4$

$y=2\left(\frac{1}{8}\right)-3\left(\frac{1}{4}\right)+4$

$y=\frac{7}{2}$

The point is then $\left(\frac{1}{2},\frac{7}{2}\right)$ .

Find the inflection points of the following function:

$f(x)=\frac{1}{6}x^4-x^3-28x^2+9$

$x=-3,x=-\frac{2}{3}$

Explanation

The points of inflection of a function are those at which its second derivative is equal to 0. First we find the second derivative of the function, then we set it equal to 0 and solve for the inflection points:

$f(x)=\frac{1}{6}x^4-x^3-28x^2+9$

$f'(x)=\frac{2}{3}x^3-3x^2-56x$

Which of the following is an -coordinate of an inflection point of the graph of the following function?

$f(x)=\frac{1}{12}x^4-2x^3+16x^2-19x+36$

$-\frac{2}{3}$

Explanation

The inflection points of a function are the points where the concavity changes, either from opening upwards to opening downwards or vice versa. The inflection points occur at the x-values where the second derivative is either zero or undefined. That means we need to find our second derivative.

We start by using the Power Rule to find the first derivative.

$f'(x)=\frac{1}{3}x^3-6x^2+32x-19$

Taking the derivative once more gives the second derivative.

We then set this derivative equal to zero and solve.

This factors nicely.

Therefore our second derivative is zero when

8 is the only one of these two amongst our choices and is therefore our answer.

Find the points of inflection of the following function:

$f(x)=\frac{1}{4}x^4-5x^3+\frac{63}{2}x^2-21$

Explanation

$f(x)=\frac{1}{4}x^4-5x^3+\frac{63}{2}x^2-21$

Determine the points of inflection of the following function:

$f(x)=\frac{1}{12}x^4+\frac{1}{3}x^3-\frac{15}{2}x^2+17$

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

Explanation

$f(x)=\frac{1}{12}x^4+\frac{1}{3}x^3-\frac{15}{2}x^2+17$

$f'(x)=\frac{1}{3}x^3+x^2-15x$

Find the x-coordinates of all points of inflection of the function .

$-\frac{1}{2}\pm\frac{\sqrt{3}}{2}$

$-\frac{1}{3}\pm\frac{\sqrt{2}}{3}$

There are no points of inflection

$0,-\frac{3}{4}\pm\frac{\sqrt{33}}{4}$

Explanation

We set the second derivative of the function equal to zero to find the x-coordinates of any points of inflection.

, and the quadratic formula yields

$x=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}$ .

Find the point(s) of inflection of the following function:

Explanation

To solve, simply differentiate twice, find when the function is equal to 0, and then plug into the first equation.

For this particular function, use to power rule to differentiate.

The power rule states,

$(x^n)'=nx^{n-1}$ .

Also recall that the derivative of a constant is zero.

Applying the power rule and rule of a constant once to the function we can find the first derivative.

Thus,

$f'(x)=3\cdot 2x^{3-1}-1\cdot 4x^{1-1}+0$

From here apply the power rule and rule of a constant once more to find the second derivative of the function.

$f''(x)=2\cdot 6x^{2-1}-0$

Now to solve for the inflection point, set the second derivative equal to zero.

$12x=0\Rightarrow x=0$

From here plug this x value into the original function to find the y value of the inflection point.

Thus, our point is (0,1).

Determine the point(s) of inflection of $y=\frac{1}{x+4}$ .

No points of inflection exist.

and

$\left(0,\frac{1}{4}\right)$

Explanation

The points of inflection exist where the second derivative is zero.

$y=\frac{1}{x+4}$

$y'=\frac{-1}{(x+4)^2}$

$y''=\frac{2}{(x+4)^3}$ which can never be . Therefore, there are no points of inflection.

Determine Points of Inflection

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