Other Points - AP Calculus AB

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Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

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Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

Compare your answer with the correct one above

Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left \{ {{-5, -1, 0, 3}} \right \}$

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Question

Find the critical points (rounded to two decimal places):

Answer

To find the critical points, set and solve for .

Differentiate:

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

Set equal to zero:

Solve for using the quadratic formula: $x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=3,\ b=6,\ c=1$

$x= \frac{-6\pm \sqrt{(6)^2-4(3)(1)}}{2(3)}$

$x= \frac{-6\pm \sqrt{36-12}}{6}}$

$x= \frac{-6+\sqrt{24}}{6}}\ and\ x= \frac{-6-\sqrt{24}}{6}}$

$c=-0.18\ and\ c=-1.82$

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Question

Find the value(s) of the critical point(s) of

.

Answer

In order to find the critical points, we must find $\ f'(x)$ and solve for $\ x$

$\ \ f'(x)=3x^2+4x+1$

Set

$\ \ 3x^2+4x+1=0$

Use the quadratic equation to solve for $\ x$ .

Remember that the quadratic equation is as follows.

$\ \ x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$ , where a,b and c refer to the coefficents in the

equation .

In this case, , , and .

After plugging in those values, we get

$\ \ x=\frac{-4\pm\sqrt{4^2-4\cdot3\cdot1}}{2\cdot3} =\frac{-4 \pm \sqrt{4}}{6}=\frac{-4 \pm 2}{6}$ .

So the critical points values are:

$\ \ c_1=\frac{-4+2}{6}=\frac{-1}{3}$

$\ \ c_2=\frac{-4-2}{6}=-1$

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Question

Find the value(s) of the critical point(s) of

.

Answer

In order to find the critical points, we must find $\ f'(x)$ and solve for $\ x$ .

$\ \ f(x)=5x^3+10x^2+x+100$

$\ \ f'(x)=15x^2+20x+1$

Set $\ f'(x)=0$

$\ \ 15x^2+20x+1=0$

Use the quadratic equation to solve for $\ x$ .

Remember that the quadratic equation is as follows.

$\ \ x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$ , where a,b and c refer to the coefficents in the equation .

In this case, , , and .

After plugging in those values, we get.

${}\ x=\frac{-20\pm\sqrt{20^2-4\cdot15\cdot1}}{2\cdot15} =\frac{-20 \pm \sqrt{340}}{30} \ \ x=\frac{-20 \pm 2\cdot\sqrt{85}}{30}$

So the critical points values are,

${}\ \ c_1=\frac{-20+2\cdot \sqrt{85}}{30}=\frac{-10+\sqrt{85}}{15} \ \ c_2=\frac{-20-2\cdot \sqrt{85}}{30}=\frac{-10-\sqrt{85}}{15}$

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Question

Find the critical points of

Answer

First we need to find .

Now we set

Now we can use the quadratic equation in order to find the critical points.

Remember that the quadratic equation is

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ,

where a,b,c refer to the coefficients in the equation

In this case, a=3, b=6, and c=1.

$x=\frac{-6\pm\sqrt{(-6)^2-4(3)(1)}}{(2)(3)}$

$x=\frac{-6\pm\sqrt{24}}{6}=\frac{-6\pm2\sqrt{6}}{6}$

Thus are critical points are

$c_1=\frac{-6+2\sqrt{6}}{6}=-1+\frac{\sqrt{6}}{3}$

$c_2=\frac{-6-2\sqrt{6}}{6}=-1-\frac{\sqrt{6}}{3}$

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Question

Find the critical points of

.

Answer

In order to find the critical points, we need to find using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Now we set , and solve for .

Thus is a critical point.

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Question

Find the critical point(s) of $f(x)=2x^{2}-5x+4$ .

Answer

To find the critical point(s) of a function , take its derivative , set it equal to , and solve for .

Given $f(x)=2x^{2}-5x+4$ , use the power rule

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ to find the derivative. Thus the derivative is, .

Since :

$x=\frac{5}{4}$

The critical point is $c=\frac{5}{4}$ .

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