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Calculus 1 : Other Points

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Example Questions

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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Example Question #12 : Other Points

Find

$\lim_{x\rightarrow 3} \frac{x^{2}+2x-15}{x-3}$ .

Possible Answers:

Limit does not exist.

Correct answer:

Explanation:

To evaluate this limit, all we need to do is factor the numerator and then cancel out the factor that is in common using the following formula:

$x^{2}+bx+c=(x+d)(x+e)$

With a simple algebra trick, we will be able to easily plug in for :

$\lim_{x\rightarrow 3} \frac{x^{2}+2x-15}{x-3}=\lim_{x\rightarrow 3}\frac{(x+5)(x-3)}{x-3}$

$=\lim_{x\rightarrow 3} x+5=3+5=8$

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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 : How To Graph Functions Of Points

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

Possible Answers:

None of the other answers.

5, -5

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 #1564 : 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 #15 : How To Graph Functions Of Points

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

Possible Answers:

-2, 1

-2, -1

1, 2

None of the other answers.

-1, 2

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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Calculus 1 : Other Points

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2581 : Calculus

Example Question #12 : Other Points

Example Question #1561 : Functions

Example Question #11 : How To Graph Functions Of Points

Example Question #2591 : Calculus

Example Question #11 : How To Graph Functions Of Points

Example Question #1564 : Functions

Example Question #1565 : Functions

Example Question #15 : How To Graph Functions Of Points

All Calculus 1 Resources

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