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Calculus 1 : Functions

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Example Questions

Example Question #1561 : Functions

Find all points on the graph of $f(x)=\cos^{2}(x)$ where the tangent line is horizontal.

Possible Answers:

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

The tangent line is never horizontal for this graph.

$x=\frac{n\pi}{3}$

$x=\frac{3n\pi}{2}$

$x=\frac{n\pi}{2}$

Correct answer:

$x=\frac{n\pi}{2}$

Explanation:

To solve this problem, we need the chain rule, the derivative of the trigonometric function cosine, and the power rule.

First let's apply the chain rule, which states:

$\frac{d}{dx} h(i(x))=h'(i(x))i'(x)$

In this problem, $h(x)=\cos^{2}(x)$ and $i(x)=\cos(x)$ .

To find h'(x) , we need the power rule which states:

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

$h'(x)=2\cos(x)^{2-1}=2\cos(x)$

To find i'(x) , we need the derivative of cosine which states:

$\frac{d}{dx} \cos(x)= -\sin(x)$

$i'(x)=-\sin(x)$

Plugging these equations into the chain rule we obtain:

$f'(x)=2\cos(x)*-\sin(x)$

To find all points where the tangent line is horizontal, we must first take the derivative of the function and then set it equal to zero:

Setting this equal to zero, we obtain:

$0=2\cos(x)*-\sin(x)$

$0=\cos(x)\sin(x)$

Therefore, either $\cos(x)=0$ or $\sin(x)=0$

Recall that from the unit circle, cosine equals zero at $\frac{\pi}{2}, \frac{3\pi}{2}$ and sine equals zero at $0, \pi, 2\pi$ .

So, at every multiple of $\frac{\pi}{2}$ , either $\cos(x)=0$ or $\sin(x)=0$ .

Therefore, $x=\frac{n\pi}{2}$ because at each multiple of $\frac{\pi}{2}$ , either $\cos(x)=0$ or $\sin(x)=0$

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Example Question #11 : How To Graph Functions Of Points

Evaluate the limit:

$\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos(x)-\sin(x)}{2\cos(2x)}$

Possible Answers:

$\frac{\sqrt2}{4}$

$\frac{\sqrt2}{2}$

$\frac{3\sqrt2}{4}$

$\frac{\sqrt3}{4}$

The limit does not exist at this point.

Correct answer:

$\frac{\sqrt2}{4}$

Explanation:

To begin, we need L'Hopital's Rule for this problem which states that if you get $\frac{0}{0}$ when you plug in the value into your function when evalutating the limit, you should take the derivative of both the numerator and the denominator and then try plugging in your value again.

$\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos(x)-\sin(x)}{2\cos(2x)}=\frac{\cos(\frac{\pi}{4})-\sin(\frac{\pi}{4})}{2\cos(2*\frac{\pi}{4})}=\frac{0}{0}$

Since this is the case, we will take the derivative of the numerator and denominator.

To take the derivative of the numerator, we need the differentiation formulas for the trigonometric functions cosine and sine.

$\frac{d}{dx} \sin(x)= \cos(x), \frac{d}{dx} \cos(x)= -\sin(x)$

So, the derivative of the numerator is $-\sin(x)-\cos(x)$

To find the derivative of the denominator, we again need the differentiation formula for cosine, as well as the chain rule.

$\frac{d}{dx} f(g(x))=f'(g(x))g'(x)$

In this problem, $f(x)=\cos(2x)$ and g(x)=2x

$\frac{d}{dx}\cos(2x)=-\sin(2x)$

$\frac{d}{dx}2x=2$

So, plugging these into the chain rule, we obtain:

$-2\sin(2x)$

Now let's put these expressions back into the numerator and denominator and again try to plug in our limit value:

$\lim_{x\rightarrow \frac{\pi}{4}}\frac{-\sin(x)-\cos(x)}{-4\sin(2x)}$

$=\frac{-\sin(\frac{\pi}{4})-\cos(\frac{\pi}{4})}{-4\sin(2*\frac{\pi}{4})}=\frac{\frac{\sqrt2}{2}+\frac{\sqrt2}{2}}{4}=\frac{\sqrt2}{4}$

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Example Question #2591 : Calculus

Evaluate the limit:

$\lim_{x\rightarrow2}\frac{\ln(x-1)}{x-2}$

Possible Answers:

The limit does not exist.

-2.2

Correct answer:

Explanation:

To solve this problem, we need L'Hopital's Rule, the derivative of the natural logarithm, the chain rule, the power rule, and the derivative of a constant.

Notice that if we plug in our value into the function, we obtain a value of $\frac{0}{0}$ .

$\frac{\ln(2-1)}{2-2}=\frac{\ln(1)}{0}=\frac{0}{0}$

L'Hopital's Rule, which states that if you plug in your limit value and obtain $\frac{0}{0}$ , you should take the derivative of the numerator and denominator and try plugging in your limit value again.

So we will take the derivative of the numerator and denominator.

For the numerator, we need the chain rule,the derivative of the natural logarithm, the derivative of a constant, and the power rule, which state:

$\frac{d}{dx}(c)=0, \frac{d}{dx} f(g(x))=f'(g(x))g'(x), \frac{d}{dx} \ln(x)=\frac{1}{x}$

For the numerator, $f(x)=\ln(x-1)$ and g(x)=x-1 .

Applying the chain rule to this expression yields:

$\frac{d}{dx} \ln(x-1)=\frac{1}{x-1}*(1x^{1-1}-0)=\frac{1}{x-1}$

To find the derivative of the denominator, we need the power rule and the derivative of a constant.

$\frac{d}{dx} x-2=1x^{1-1}-0=1$

So now we have:

$\lim_{x\rightarrow2}\frac{\frac{1}{x-1}}{1}=\lim_{x->2}\frac{1}{x-1}=\frac{1}{2-1}=\frac{1}{1}=1$

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Example Question #11 : Other Points

Find the x-coordinate of the critical points of x^3-15x^2+75x-45 .

Possible Answers:

5, -5

None of the other answers.

Correct answer:

Explanation:

We need to differentiate term by term, applying the power rule,

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

This gives us

$\begin{align*} (x^3-15x^2+75x-45)' &= 3x^2-30x+75 \end{align*}$

The x-coordinate of the critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:

$\begin{align*} x &= \frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}\\ &=\frac{30\pm\sqrt{900-4\cdot 3\cdot 75}}{2\cdot 3} \\ &=\frac{30\pm\sqrt0}{6}\\ &=5 \end{align*}$

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Example Question #1561 : Functions

Find the x value of the critical points of x^3-3x^2-2 .

Possible Answers:

0, 2

0, -2

Correct answer:

0, 2

Explanation:

We need to differentiate term by term, applying the power rule,

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

This gives us

$\begin{align*} (x^3-3x^2-2)' &= 3x^2-6x \end{align*}$

The critical points are the points where the derivative equals 0. To find those x values, we can use the quadratic formula:

$\begin{align*} x &= \frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a} \\ &= \frac{6\pm\sqrt{36-4\cdot 3\cdot 0}}{6}\\ &=\frac{0}{6} \text{ or } \frac{12}{6} \\ &=0 \text{ or } 2 \end{align*}$

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Example Question #1565 : Functions

Find the x values of the critical points of 2x^3-6x+2 .

Possible Answers:

None of the other answers.

-1, 1

Correct answer:

-1, 1

Explanation:

We need to differentiate term by term, applying the power rule,

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

This gives us

$\begin{align*} (2x^3-6x+2)' &= 6x^2-6 \end{align*}$

The critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:

$\begin{align*} x &= \frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}\\ &=\frac{0\pm\sqrt{0-4\cdot 6\cdot -6}}{2\cdot 6} \\ &=\frac{0\pm\sqrt{144}}{12}\\ &=\pm1 \end{align*}$

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Example Question #41 : Points

Find the x values for critical points of 2x^3+3x^2-12x+5 .

Possible Answers:

-2, 1

None of the other answers.

1, 2

-1, 2

-2, -1

Correct answer:

-2, 1

Explanation:

We need to differentiate term by term, applying the power rule,

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

This gives us

$\begin{align*} (2x^3+3x^2-12x+5)' &= 6x^2+6x-12 \end{align*}$

The critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:

$\begin{align*} x &= \frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a} \\ &= \frac{-6\pm\sqrt{36-4\cdot 6\cdot -12}}{12}\\ &= \frac{-6\pm\sqrt{384}}{12}\\ &= \frac{-6\pm18}{12}\\ &=\frac{12}{12} \text{ or } \frac{-24}{12} \\ &=1 \text{ or } -2 \end{align*}$

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Example Question #1 : Decreasing Intervals

Find the interval(s) where the following function is decreasing. Graph to double check your answer.

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

Possible Answers:

Never

$(-\infty,-5) \cup (1,\infty)$

Always

(-5,1)

Correct answer:

(-5,1)

Explanation:

To find when a function is decreasing, you must first take the derivative, then set it equal to 0, and then find between which zero values the function is negative.

First, take the derivative:

y'=x^2+4x-5

Set equal to 0 and solve:

x^2+4x-5=0

(x+5)(x-1)=0

x=-5,1

Now test values on all sides of these to find when the function is negative, and therefore decreasing. I will test the values of -6, 0, and 2.

y'=x^2+4x-5

y'(-6)=(-6)^2+4(-6)-5=7

y'(0)=(0)^2+4(0)-5=-5

y'(2)=(2)^2+4(2)-5=7

Since the only value that is negative is when x=0, the interval is only decreasing on the interval that includes 0. Therefore, our answer is:

(-5,1)

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Example Question #2 : Decreasing Intervals

Find the interval(s) where the following function is decreasing. Graph to double check your answer.

$y=\frac{1}{3}x^3-5x^2+9x+5$

Possible Answers:

Never

Always

(1,9)

$(-\infty,1)\cup (9,\infty)$

Correct answer:

(1,9)

Explanation:

To find when a function is decreasing, you must first take the derivative, then set it equal to 0, and then find between which zero values the function is negative.

First, take the derivative:

y'=x^2-10x+9

Set equal to 0 and solve:

x^2-10x+9=0

(x-9)(x-1)=0

x=1,9

Now test values on all sides of these to find when the function is negative, and therefore decreasing. I will test the values of 0, 2, and 10.

y'=x^2-10x+9

y'(0)=(0)^2-10(0)+9=9

y'(2)=(2)^2-10(2)+9=-7

y'(10)=(10)^2-10(10)+9=9

Since the only value that is negative is when x=0, the interval is only decreasing on the interval that includes 2. Therefore, our answer is:

(1,9)

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Example Question #1 : How To Find Decreasing Intervals By Graphing Functions

Is f(x) increasing or decreasing on the interval [4,7] ?

f(x)=-5x^4+6x^2-5x+3

Possible Answers:

Increasing, because the first derivative is positive on the interval [4,7] .

The function is neither increasing nor decreasing on the interval [4,7] .

Decreasing, because the first derivative of f'(x) is negative on the function [4,7] .

Increasing because the second derivative is positive on the interval [4,7] .

Decreasing, because the first derivative is positive on the interval [4,7] .

Correct answer:

Decreasing, because the first derivative of f'(x) is negative on the function [4,7] .

Explanation:

To find the an increasing or decreasing interval, we need to find out if the first derivative is positive or negative on the given interval. So, find f'(x) by decreasing each exponent by one and multiplying by the original number.

f(x)=-5x^4+6x^2-5x+3

f'(x)=-20x^3+12x-5

Next, we can find f'(4) and f'(7) and see if they are positive or negative.

$f'(4)=-20\cdot4^3+12\cdot4-5=-1237$

$f'(7)=-20\cdot7^3+12\cdot7-5=-6781$

Both are negative, so the slope of the line tangent to f(x) is negative, so f(x) is decreasing.

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1561 : Functions

Example Question #11 : How To Graph Functions Of Points

Example Question #2591 : Calculus

Example Question #11 : Other Points

Example Question #1561 : Functions

Example Question #1565 : Functions

Example Question #41 : Points

Example Question #1 : Decreasing Intervals

Example Question #2 : Decreasing Intervals

Example Question #1 : How To Find Decreasing Intervals By Graphing Functions

All Calculus 1 Resources

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