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Calculus 1 : Points

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Example Question #12 : Other Points

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 #12 : How To Graph Functions Of 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 #13 : How To Graph Functions Of Points

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 #14 : How To Graph Functions Of Points

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 : Points

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #12 : Other Points

Example Question #12 : How To Graph Functions Of Points

Example Question #13 : How To Graph Functions Of Points

Example Question #14 : How To Graph Functions Of Points

Example Question #15 : How To Graph Functions Of Points

All Calculus 1 Resources

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