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Calculus 1 : Graphing Functions

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Example Questions

Example Question #2 : How To Graph Functions Of Points

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

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

Possible Answers:

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

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

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

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

There are no real answers.

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 : How To Graph Functions Of Points

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

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

Possible Answers:

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

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

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

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

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

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:

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

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

The critical points are complex.

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

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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Example Question #5 : Other Points

Find the critical points of

f(x)=x^2-2x+1 .

Possible Answers:

There are no critical points.

x=2

x=1

x=0

Correct answer:

x=1

Explanation:

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

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

f'(x)=2x-2

Now we set f'(x)=0 , and solve for .

2x-2=0

2x=2

x=1

Thus x=1 is a critical point.

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Example Question #2 : How To Graph Functions Of Points

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

Possible Answers:

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

$c=\frac{4}{5}$ and $c=-\frac{4}{5}$

$c=\frac{5}{4}$ and $c=-\frac{5}{4}$

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

$c=\frac{5}{4}$ and $c=-\frac{4}{5}$

Correct answer:

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

Explanation:

To find the critical point(s) of a function f(x) , take its derivative f'(x) , 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, f'(x)=4x-5=0 .

Since 4x-5=0 :

4x=5

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

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

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Example Question #2 : Other Points

Find the critical points of

f(x)=3x^3-20x^2+4x+99 .

Possible Answers:

x=0

$c=\frac{20-2\sqrt{91}}{9}$

$\\ c_1=\frac{20+2\sqrt{91}}{9} \\ \\ c_2=\frac{20-2\sqrt{91}}{9}$

There are no critical points

$\\ c=\frac{20+2\sqrt{91}}{9}$

Correct answer:

$\\ c_1=\frac{20+2\sqrt{91}}{9} \\ \\ c_2=\frac{20-2\sqrt{91}}{9}$

Explanation:

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

f(x)=3x^3-20x^2+4x+99

f'(x)=9x^2-40x+4 .

Now we set f'(x)=0 .

9x^2-40x+4=0

Now we use the quadratic equation in order to solve for .

Remember that the quadratic equation is as follows.

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

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

ax^2+bx+c=0 .

In this case, a=9, b=-40, c=4.

$x=\frac{40\pm\sqrt{(-40)^2-4(9)(4)}}{2(9)}$

$x=\frac{40\pm\sqrt{1600-144}}{18}$

$x=\frac{40\pm\sqrt{1456}}{18}=\frac{40\pm4\sqrt{91}}{18}$

Then are critical points are:

$c_1=\frac{40+4\sqrt{91}}{18}=\frac{20+2\sqrt{91}}{9}$

$c_2=\frac{40-4\sqrt{91}}{18}=\frac{20-2\sqrt{91}}{9}$

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Example Question #2 : Other Points

Find all the critical points of

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

Possible Answers:

x=0

There are no critical points.

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

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

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

Correct answer:

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

Explanation:

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

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

f'(x)=3x^2-2

Now we set f'(x)=0 .

3x^2-2=0

3x^2=2

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

$x=\pm\sqrt{\frac{2}{3}}$

Thus the critical points are at

$x=\sqrt{\frac{2}{3}}$ , and

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

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Example Question #2 : Other Points

Find the critical points of the following function:

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

Possible Answers:

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

$x=\left \{ {0, 1.332, 6.013} \right \}$

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

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

$x=\left \{ {-6.013, -1.332, 0} \right \}$

Correct answer:

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

Explanation:

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

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

y'=x^5+3x^4-13x^3-15x^3

Critical points occur where the derivative equals zero.

0=x^5+3x^4-13x^3-15x^3

0=x^2(x+1)(x+5)(x-3)

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

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Example Question #3 : Other Points

Determine the point on the graph that is not changing if $f(x)=x^{2}-12x+32$ .

Possible Answers:

(6,-2)

(3,-2)

(0,-4)

(6,-4)

(0,0)

Correct answer:

(6,-4)

Explanation:

To find the point where the graph of f(x) is not changing, we must set the first derivative equal to zero and solve for .

To evaluate this derivate, we need the following formulae:

$\frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1, \frac{d}{dx}cx=c$

$\frac{d}{dx}(x^{n}))=nx^{n-1}, \frac{d}{dx}(cx^{n})=ncx^{n-1}$

$f'(x)=2x^{2-1}-12x^{1-1}+0=2x-12$

Now, setting the derivate equal to to find where the graph is not changing:

2x-12=0 ==> x=6

Now, to find the corresponding value, we plug this value back into f(x) :

$f(6)=6^{2}-12(6)+32=36-72+32=-4$

Therefore, the point where f(x) is not changing is (6,-4)

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Example Question #2581 : Calculus

Find the limit: $\lim_{x\rightarrow 0} \frac{\sin(6x)}{x}$

Possible Answers:

Limit does not exist.

Correct answer:

Explanation:

To evaluate this limit, we must use L'Hopital's Rule:

If $\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\frac{0}{0}$ , take the derivative of both f(x) and g(x) and then plug in to obtain $\frac{f'(c)}{g'(c)}$

We will also need the power rule, the derivative of the trigonometric function sine, and the chain rule.

Since when we plug in in the numerator and denominator, we obtain a result of $\frac{0}{0}$ , we can use L'Hopitals rule.

To take the derivative of the numerator we need the chain rule, the derivative of the trigonometric function sine, and the power rule.

$\frac{d}{dx}(ax^{n})=anx^{n-1}, \frac{d}{dx} f(g(x))=f'(g(x))g'(x), \frac{d}{dx} \sin(x)= \cos(x)$

Applying the chain rule to the numerator with $f(x)=\sin(6x)$ and g(x)=6x , we see that:

$f'(x)=\cos(6x)$ and $g'(x)=6x^{1-1}=6$ .

Now plugging these into the chain rule, we obtain:

$6\cos(6x)$

Now, to find the derivative of the denominator, we need the power rule again:

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

Now that we have found the derivative of the numerator and denominator, we can apply L'Hopital's Rule:

$\lim_{x\rightarrow 0} \frac{6\cos(6x)}{1}=6\cos(0)=6$

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Calculus 1 : Graphing Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2 : How To Graph Functions Of Points

Example Question #1 : How To Graph Functions Of Points

Example Question #2 : Other Points

Example Question #5 : Other Points

Example Question #2 : How To Graph Functions Of Points

Example Question #2 : Other Points

Example Question #2 : Other Points

Example Question #2 : Other Points

Example Question #3 : Other Points

Example Question #2581 : Calculus

All Calculus 1 Resources

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