Derivatives - Pre-Calculus

Card 0 of 20

Question

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

$f(x)=\frac{1}{12}x^4-\frac{1}{2}x^3-5x^2+4x+17$

Answer

The points of inflection of a function are the points at which its concavity changes. The concavity of a function is described by its second derivative, which will be equal to zero at the inflection points, so we'll start by finding the first derivative of the function:

$f(x)=\frac{1}{12}x^4-\frac{1}{2}x^3-5x^2+4x+17$

$f'(x)=\frac{1}{3}x^3-\frac{3}{2}x^2-10x+4$

Next we'll take the derivative one more time to get the second derivative of the original function:

$f'(x)=\frac{1}{3}x^3-\frac{3}{2}x^2-10x+4$

Now that we have the second derivative of the function, we can set it equal to 0 and solve for the points of inflection:

$x^2-3x-10=0\rightarrow (x-5)(x+2)=0\rightarrow x=-2,x=5$

Compare your answer with the correct one above

Question

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

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

Answer

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.

Compare your answer with the correct one above

Question

Determine the points of inflection of the following function:

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

Answer

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:

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

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

Compare your answer with the correct one above

Question

Find the inflection points of the following function:

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

Answer

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:

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

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

Compare your answer with the correct one above

Question

Find the points of inflection of the following function:

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

Answer

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:

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

Compare your answer with the correct one above

Question

Determine the points of inflection, if any, of the following function:

Answer

The points of inflection of a function are those at which its concavity changes. The concavity of a function is described by its second derivative, and when the second derivative is 0 a point of inflection occurs. We find the second derivative of the function and then set it equal to 0 to solve for the inflection points:

$6x-10=0\rightarrow 6x=10\rightarrow x=\frac{5}{3}$

So the function has only one point of inflection at x=5/3.

Compare your answer with the correct one above

Question

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

Answer

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

, and we set this equal to zero.

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

Compare your answer with the correct one above

Question

Find the point of inflection of the function .

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

Determine the x-coordinate(s) of the point(s) of inflection of the function $y=\frac{1}{6}x^4-\frac{2}{3}x^3-15x^2+12x+1$ .

Answer

Any points of inflection that exist will be found where the second derivative is equal to zero.

$y=\frac{1}{6}x^4-\frac{2}{3}x^3-15x^2+12x+1$

$y'=\frac{2}{3}x^3-2x^2-30x+12$

.

Since , we can focus on . Thus

, and .

Compare your answer with the correct one above

Question

Find the point(s) of inflection of $y=(x+3)^\frac{2}{3}$ .

Answer

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

$y=(x+3)^\frac{2}{3}$

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

$y''=-\frac{2}{3}(x+3)^\frac{-4}{3}$

$0=-\frac{2}{3}(x+3)^\frac{-4}{3}$

$0=-\frac{2}{3}\cdot\frac{1}{\sqrt[3]{x+3}^4}$ .

This will never be , so there are no points of inflection.

Compare your answer with the correct one above

Question

List the interval(s) where the function is concave up.

Answer

The graph is concave up where the second derivative is positive. Let us first find out if there are any points of inflection to narrow our search.

$x=\frac{2}{3}$ .

Now we can perform the second derivative concavity test on points on either side of $x=\frac{2}{3}$ . Let us try .

The second derivative at gives us which is less than zero, so the graph is concave down in this interval. The second derivative at gives us , which is positive. Hence, the graph is concave up on the interval $\left(\frac{2}{3},\infty\right)$ .

Compare your answer with the correct one above

Question

Find the x-coordinate(s) of the point(s) of inflection of $y=\frac{x^3}{x}$ .

Answer

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

$y=\frac{x^3}{x}=x^2$

Now therefore, there are no points of inflection.

Compare your answer with the correct one above

Question

Find the inflection point(s) of $y=\frac{1}{12}x^4-\frac{1}{6}x^3-x^2+2x+3$ .

Answer

The points of inflection, if any exist, will be found where the second derivative is zero.

$y=\frac{1}{12}x^4-\frac{1}{6}x^3-x^2+2x+3$

$y'=\frac{1}{3}x^3-\frac{1}{2}x^2-2x+2$

.

To find the y-coordinates, we simply plug the x-coordinates in to the original function.

$y=\frac{1}{12}(-1)^4-\frac{1}{6}(-1)^3-(-1)^2+2(-1)+3=\frac{1}{4}$ .

So $(-1,\frac{1}{4})$ is an inflection point.

Also, $y=\frac{1}{12}(2)^4-\frac{1}{6}(2)^3-(2)^2+2(2)+3=3$ , so is another inflection point.

Compare your answer with the correct one above

Question

Find the x-coordinate(s) of the point(s) of inflection of $y=\frac{1}{6}x^4-x^2-3x+1$ .

Answer

The inflection points, if they exist, will occur where the second derivative is zero.

$y=\frac{1}{6}x^4-x^2-3x+1$

$y'=\frac{2}{3}x^3-2x-3$

$x=\pm1$

Compare your answer with the correct one above

Question

Find the point(s) of inflection of the function .

Answer

The point of inflection will exist where the second derivative equals zero.

.

Now we need the y-coordinate of the point.

Thus the inflection point is at .

Compare your answer with the correct one above

Question

List the intervals and determine where the graph of $y=\frac{1}{3}x^4+\frac{1}{3}x^3-6x^2+x+5$ is concave up and concave down.

Answer

We need to set the second derivative equal to zero to determine where the inflection points are.

$y=\frac{1}{3}x^4+\frac{1}{3}x^3-6x^2+x+5$

$y'=\frac{4}{3}x^3+x^2-12x+1$

$x=-2,\frac{3}{2}$ are the x-coordinates of our inflection points. Thus the intervals of concavity are $(-\infty,-2)$ , $\left(-2,\frac{3}{2}\right)$ , and $\left(\frac{3}{2},\infty\right)$ . We can use as our test points.

, so the graph is concave up on $(-\infty,-2)$ .

, so the graph is concave down on $\left(-2,\frac{3}{2}\right)$ .

, so the graph is concave up on $\left(\frac{3}{2},\infty\right)$ .

Compare your answer with the correct one above

Question

Determine the points of inflection for the following function.

Answer

To find inflection points, take the second derivative and set it equal to .

Since is never , this function does not have any inflection points. Thus, the answer is none.

Compare your answer with the correct one above

Question

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

Answer

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).

Compare your answer with the correct one above

Tap the card to reveal the answer