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Calculus 1 : Points

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Example Question #2572 : Calculus

Where does the function $f(x)=5x^{3}+\frac{5}{3}x^2+11$ have a point of inflection?

Possible Answers:

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

$x=\frac{1}{9}$

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

$x=-\frac{1}{9}$

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

Correct answer:

$x=-\frac{1}{9}$

Explanation:

The point of inflection of an object can be found by setting the second derivative of the function equal to zero and solving.

$f'(x)=15x^2+\frac{10}{3}x.$

$f''(x)=30x+\frac{10}{3}=0.$

$30x=-\frac{10}{3}, x=-\frac{1}{9}.$

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Example Question #2573 : Calculus

Find the value of the point of inflection of the function $f(x)=\frac{12x-\frac{8}{3}x^5}{x^2}$ .

Possible Answers:

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

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

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

$x=(\frac{2}{3})^{\frac{1}{4}}$

$x=(\frac{3}{2})^{\frac{1}{4}}$

Correct answer:

$x=(\frac{3}{2})^{\frac{1}{4}}$

Explanation:

First, we need to simplify the expression. Sine the numerator has more terms than the denominator, we can simply divide out both terms.

$\frac{12x-\frac{8}{3}x^5}{x^2}=12x^{-1}-\frac{8}{3}x^3$ .

Now, we must compute the second derivative of this function, and set it equal to zero.

$f'(x)=-12x^{-2}-8x^2$ .

$f''(x)=24x^{-3}-16x=0.$

$x^{-4}=\frac{2}{3}$ .

$x=(\frac{3}{2})^{\frac{1}{4}}$ .

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Example Question #2574 : Calculus

Find the points of inflection for the following function:

f(x)=x^3+5x^2+5x

Possible Answers:

$(-\frac{5}{3}, \frac{25}{27})$

$(-\frac{5}{3}, -\frac{25}{27})$

(0, 0)

$(-\frac{5}{3}, 0)$

Correct answer:

$(-\frac{5}{3}, \frac{25}{27})$

Explanation:

To find the points of inflection, we must find the points at which the second derivative of the function changes sign.

First, we must find the second derivative:

f'(x)=3x^2+10x+5

f''(x)=6x+10

The derivatives were found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, we must find the x value at which the second derivative is equal to zero:

$x=-\frac{5}{3}$

Using this value, we make our intervals on which we examine the sign of the second derivative:

$(-\infty, -\frac{5}{3}), (-\frac{5}{3}, \infty)$

Note that at the endpoints of the intervals, the second derivative is neither positive nor negative.

Now, simply plug in any value on the interval into the second derivative function and check the sign. On the first interval, the second derivative is negative, while on the second interval, it is positive. Thus, a point of inflection exists at $x=-\frac{5}{3}$ . To get the y-coordinate, simply evaluate the function at this x value:

$f(-\frac{5}{3})=\frac{25}{27}$

The point of inflection is $(-\frac{5}{3}, \frac{25}{27})$ .

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Example Question #2575 : Calculus

Determine the points of inflection for the following function:

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

Possible Answers:

x=2

x=-1

x=0

x=1

Correct answer:

x=-1

Explanation:

To determine the points of inflection, we must determine the x values at which the second derivative changes in sign.

So, we must first find the second derivative:

f'(x)=x^2+2x+1

f''(x)=2x+2

The derivatives were found using the rule

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, we must find the values at which the second derivative is equal to zero:

2(x+1)=0

x=-1

Now, using this zero, we can make the intervals on which we check the sign of the second derivative:

$(-\infty, -1), (-1, \infty)$

Note that at the endpoints, the second derivative is neither positive nor negative.

Plug in any value on each interval into the second derivative function and check the sign. On the first interval, the second derivative is negative, while on the second, it is positive. Because a sign change occured at x=-1 , there exists the point of inflection.

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Example Question #2576 : Calculus

Determine the points of inflection for the following function:

f(x)=x^3+10x^2+x

Possible Answers:

$x=\frac{-20+\sqrt{388}}{6}$

$x=-\frac{20}{3}$

$x=-\frac{10}{3}$

$x=\frac{10}{3}$

Correct answer:

$x=-\frac{10}{3}$

Explanation:

To determine the points of inflection of the function, we must determine where the second derivative of the function changes sign.

First, we find the second derivative:

f'(x)=3x^2+20x+1

f''(x)=6x+20

The derivatives were found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, we must find where the second derivative equals zero:

2(3x+10)=0

$x=-\frac{10}{3}$

Using this value, we can now create the intervals on which we check the sign of the second derivative:

$(-\infty, -\frac{10}{3}), (-\frac{10}{3}, \infty)$

Note that at the endpoints of the intervals, the second derivative is neither positive nor negative.

To check the second derivative's sign on each interval, simply plug in any point on the interval into the second derivative function. On the first interval, the second derivative is negative, while on the second interval, it is positive. Because the sign of the second derivative changes at $x=-\frac{10}{3}$ , there exists the point of inflection.

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Example Question #3 : Inflection Points

What are the coordinates of the points of inflection for the graph

$y=\frac{x^4}{12}+x^3+4x^2-8x+5?$

Possible Answers:

x=4,-2

x=-4,-2

There are no points of inflection on this graph.

x=-4,2

x=4,2

Correct answer:

x=-4,-2

Explanation:

Infelction points are the points of a graph where the concavity of the graph changes. The inflection points of a graph are found by taking the double derivative of the graph equation, setting it equal to zero, then solving for .

To take the derivative of this equation, we must use the power rule,

$\frac{d}{dx}x^n=nx^{n-1}$ .

We also must remember that the derivative of an constant is 0.

After taking the first derivative of the graph equation using the power rule, the equation becomes

$y^{'}=\frac{x^3}{3}+3x^2+8x-8$ .

In this problem the double derivative of the graph equation comes out to $y^{''}=x^2+6x+8$ , factoring this equation out it becomes $y^{''}=(x+4)(x+2)$ .

Solving for when the equation is set equal to zero, the inflection points are located at x=-4 , -2 .

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Example Question #2577 : Calculus

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

f(x)= x^3+3x^2+x

Possible Answers:

$c=-0.25\ and\ c=-2.41$

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

$c=0.18\ and\ c=1.82$

$c=0.25\ and\ c=2.41$

Correct answer:

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

Explanation:

To find the critical points, set f{}'(x)=0 and solve for .

f(x)= x^3+3x^2+x

Differentiate:

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

Set equal to zero:

3x^2+6x+1=0

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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Example Question #2578 : Calculus

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

f(x)=x^3+2x^2+x+4 .

Possible Answers:

$\\ c_1=-\frac{1}{3} \\ \\ c_2=-1$

$\\ c_1=3 \\ \\ c_2=-1$

There are no real answers.

$\\ c_1=-3 \\ \\ c_2=1$

$\\ c_1=\frac{1}{3} \\ \\ c_2=1$

Correct answer:

$\\ c_1=-\frac{1}{3} \\ \\ c_2=-1$

Explanation:

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

f(x)=x^3+2x^2+x+4

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

Set f'(x)=0

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

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 ax^2+bx+c=0 .

In this case, a=3 , b=4 , and c=1 .

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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Example Question #1 : Other Points

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

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

Possible Answers:

$\\ c_1=\frac{-10+\sqrt{85}}{15} \\ \\ c_2=\frac{-10-\sqrt{85}}{15}$

$\\ c_1=\frac{-10+\sqrt{85}}{30} \\ \\ c_2=\frac{-10-\sqrt{85}}{30}$

$\\ c_1=0 \\ \\ c_2=0$

$\\ c_1=1 \\ \\ c_2=0$

$\\ c_1=\frac{\sqrt{85}}{15} \\ \\ c_2=\frac{-\sqrt{85}}{15}$

Correct answer:

$\\ c_1=\frac{-10+\sqrt{85}}{15} \\ \\ c_2=\frac{-10-\sqrt{85}}{15}$

Explanation:

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 ax^2+bx+c=0 .

In this case, a=15 , b=20 , and c=1 .

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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Example Question #2 : Other Points

Find the critical points of

f(x)=x^3+3x^2+x+5

Possible Answers:

$\\ c_1=\frac{\sqrt{6}}{3}\\\\\ c_2=-\frac{\sqrt{6}}{3}$

$\\ c_1=-1+\frac{\sqrt{6}}{3}\\\\\ c_2=-1-\frac{\sqrt{6}}{3}$

The critical points are complex.

$\\ c_1=0\\\\\ c_2=-1-\frac{\sqrt{6}}{3}$

Correct answer:

$\\ c_1=-1+\frac{\sqrt{6}}{3}\\\\\ c_2=-1-\frac{\sqrt{6}}{3}$

Explanation:

First we need to find f'(x) .

f(x)=x^3+3x^2+x+5

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

Now we set f'(x)=0.

3x^2+6x+4=0.

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

ax^2+bx+c=0

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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Calculus 1 : Points

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2572 : Calculus

Example Question #2573 : Calculus

Example Question #2574 : Calculus

Example Question #2575 : Calculus

Example Question #2576 : Calculus

Example Question #3 : Inflection Points

Example Question #2577 : Calculus

Example Question #2578 : Calculus

Example Question #1 : Other Points

Example Question #2 : Other Points

All Calculus 1 Resources

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