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Calculus 1 : Lines

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Example Questions

Example Question #2711 : Calculus

Find the equation of the tangent line, where

f(x)=x^4-x^2+x-5 , at x=1 .

Possible Answers:

y=3x+1

y=3x-7

y=3x-1

y=3x+7

y=x

Correct answer:

y=3x-7

Explanation:

In order to find the equation of the tangent line at x=1 , we first find the slope.

To do this we need to find f'(x) .

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

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

Since we have found f'(x) , now we simply plug in 1.

f'(1)=4(1)^3-2(1)+1=3

Now we need to plug in 1, into f(x) , to find a point that the tangent line touches.

f(1)=(1)^4-(1)^2+1-5=-4

Now we can use point-slope form to figure out what the equation of the tangent line is at x=1 .

Remember that point-slope for is

y-y_0=m(x-x_0)

where x_0 and y_0 is the point where the tangent line touches f(x) , and is the slope of the tangent line.

In our case, x_0=1 , y_0=-4 , and m=3 .

y+4=3(x-1)

y+4=3x-3

Thus our tangent line equation at x=1 is

y=3x-7 .

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Example Question #1 : Equation Of Line

Find the equation of the tangent line of

f(x)=x^3+x , at x=1 .

Possible Answers:

y=4x

y=4x+2

y=4x-2

y=2x+7

y=-4x-2

Correct answer:

y=4x-2

Explanation:

In order to find the equation of the tangent line at x=1 , we first find the slope.

To do this we need to find f'(x) using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

f(x)=x^3+x

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

Since we have found f'(x) , now we simply plug in 1.

f'(1)=3(1)^2+1=4

Now we need to plug in 1, into f(x) , to find a point that the tangent line touches.

f(1)=(1)^3+1=2

Now we can use point-slope form to figure out what the equation of the tangent line is at x=1 .

Remember that point-slope for is

y-y_0=m(x-x_0)

where x_0 and y_0 is the point where the tangent line touches f(x) , and is the slope of the tangent line.

In our case, x_0=1 , y_0=2 , and m=4 .

y-2=4(x-1)

y-2=4x-4

Thus our tangent line equation at x=1 is

y=4x-2 .

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Example Question #11 : How To Find Equation Of Line By Graphing Functions

Give the general equation for the line tangent to f(x) = x^2 - 8x + 4 at the point $(x_{0},y_{0})$ .

Possible Answers:

$2x_{0}^2 - 8$

$-x_{0}^2 - x(x_{0} - 8) - 4$

$x_{0}^2 - x(2x_{0} - 8) - 4$

$-x_{0}^2 + x_0(2x-8) + 4$

$2x_{0}^2 +8x_{0} + 4$

Correct answer:

$-x_{0}^2 + x_0(2x-8) + 4$

Explanation:

The equation of the tangent line has the form $y - y_{0} = m(x-x_{0})$ .

The slope can be determined by evaluating the derivative of the function at $x = x_{0}$ .

f'(x) = 2x - 8

$f'(x_{0}) = 2x_{0} - 8$

Plugging this into the point slope equation, we get

$y - y_{0} = (2x_{0} - 8)(x - x_{0})$

$y_{0}$ can be determined by evaluating the original function at $x = x_{0}$ .

$y_{0} = x_{0}^2 - 8x_0 + 4$

Plugging this into the previous equation and simplifying gives us

$-x_{0}^2 + x_0(2x-8) + 4$

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Example Question #2714 : Calculus

Find the slope of the line tangent to the following function at $\frac{\pi}{4}$ .

$\cos(2x)+\sin (x)$

Possible Answers:

$-2+\frac{\sqrt{2}}{2}$

None of these

$-2+\frac{\sqrt{3}}{2}$

$-1+\frac{\sqrt{2}}{2}$

$-2+\frac{\sqrt{2}}{4}$

Correct answer:

$-2+\frac{\sqrt{2}}{2}$

Explanation:

To find the slope of the line tangent you must take the derivative of the function. The derivative of cosine is negative sine and the derivative of sine is cosine.

This makes the derivative of the function

$-2\sin (2x)+\cos (x)$ .

Plug in the given x to get the slope.

$-2\sin \left(\frac{2\pi}{4}\right)+\cos \left(\frac{\pi}{4}\right)$

$=-2+\frac{\sqrt{2}}{2}$

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Example Question #11 : Lines

Find the slope of the tangent line to the following function at x=1 .

3x^2-6x+12-ln(x^2)

Possible Answers:

None of these

Correct answer:

Explanation:

To find the slope of the line tangent to the function at a point you must first find the derivative.

The power rule states that the derivative of x^n is $nx^{n-1}$ .

The derivative of ln(x) is $\frac{1}{x}$ .

The derivative of the function is

$6x-6-\frac{2x}{x^2}$ .

Plugging in 1 for x gives

$6(1)-6-\frac{2(1)}{1^2}=-2$ .

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Example Question #2711 : Calculus

Given the differential function f(x) , we are told that f(2)=0 , f'(2)=5 , and f''(2)=0 . Which of the following must be true?

Possible Answers:

The line 5x-y=10 is tangent to f(x) .

f(x) must have at least one relative maximum.

f(x) has a point of inflection at (2,0) .

f(x) is increasing over the interval [0,2] .

f(x) is decreasing at x=2 .

Correct answer:

The line 5x-y=10 is tangent to f(x) .

Explanation:

" f(x) is decreasing at x=2 ." is incorrect. The function is increasing at x=2 because f'(2)>0 .

" f(x) is increasing over the interval [0,2] ." is possibly true, but there is not enough information to conclude that it must be true.

" f(x) has a point of inflection at (2,0) ." is possibly true. Although we know that f''(2)=0 , a requirement for an inflection point, we do not know that f''(x) changes signs at x=2 .

" f(x) must have at least one relative maximum." is possibly true, but there is not enough information to conclude that it must be true.

"The line 5x-y=10 is tangent to f(x) ." must be true. Because f(2)=0 , the function travels through the point (2,0) . Because f'(2)=5 , the slope of the line tangent to the curve at (2,0) is 5. Use point-slope form to determine the equation of the tangent line.

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Example Question #242 : Graphing Functions

Find the equation for the line tangent to the curve $\ln(\frac{x}{2})+2x^2-2$ at (2,6) .

Possible Answers:

$f(x)=\frac{17x}{2}-11$

f(x)=17x-28

$f(x)=\frac{17x}{2}-6$

f(x)=5x-6

f(x)=5x-4

Correct answer:

$f(x)=\frac{17x}{2}-11$

Explanation:

The derivative of the function is $\frac{1}{x}+4x$ , and is found using the power rule

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

and the rule for the derivative of natural log which is,

$\frac{d}{dx}ln(x)=\frac{1}{x}$

so plugging in gives $\frac{17}{2}$ , which must be the slope of the line since the tangent line's slope is determined by the derivative.

Thus, the line is of the form $f(x)=\frac{17x}{2}+b$ , where b is unknown.

Solve for b by setting the equation equal to and plugging in for x since that is the given point.

$0=\frac{17(2)}{2}+b$ $\Rightarrow b=-11$ , which gives us $f(x)=\frac{17x}{2}-11$

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Example Question #11 : How To Find Equation Of Line By Graphing Functions

Find the slope of the tangent line through the given point of the following function.

g(x)=7x^3+21x at the point (1,28)

Possible Answers:

105

Correct answer:

Explanation:

In order to find the slope of the tangent line through a certain point, we must find the rate of change (derivative) of the function. The derivative of g(x)=7x^3+21x is written as g'(x)=21x^2+21 . This tells us what the slope of the tangent line is through any point in our function g(x) . In other words, all we need to do is plug-in (because our point (1,28) has an x-value of 1) into g'(x) . This will give us our answer, .

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Example Question #17 : Equation Of Line

Find the tangent line to the function f(x)=x^3-4x^2+5 at the point (3,-4) .

Possible Answers:

y=x-4

y=3

y=3x-13

y=3x-4

y=3x^2-8x

Correct answer:

y=3x-13

Explanation:

To find the tangent line one must first find the slope, this can be given by the derivative evaluated at a point.

To find the derivative of this function use the power rule which states,

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

The derviative of f(x)=x^3-4x^2+5 is $f'(x)=3x^{2}-8x$ .

Evaluated at our point x=3 ,

$\\f'(x)=3x^{2}-8x \\f'(3)=3(3)^2-8(3) \\f'(3)=27-24=3$

we find that the slope, m is also 3.

Now we may use the point-slope equation of a line to find the tangent line.

The point slope equation is

$y=m(x-x_{1})+y_{_{1}}$ Where $(x_{1},y_{1})$ is the point at which the line is tangent.

Using this definition we find the tangent line to be defined by y=3x-13 .

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Example Question #1 : Slope

What is the slope of the tangent line of f(x) = 3x⁴ – 5x³ – 4x at x = 40?

Possible Answers:

743,996

331,841

684,910

None of the other answers

768,000

Correct answer:

743,996

Explanation:

The first derivative is easy:

f'(x) = 12x³ – 15x² – 4

The slope of the tangent line is found by calculating f'(40) = 12 * 40³ – 15 * 40² – 4 = 768,000 – 24,000 – 4 = 743,996

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Calculus 1 : Lines

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2711 : Calculus

Example Question #1 : Equation Of Line

Example Question #11 : How To Find Equation Of Line By Graphing Functions

Example Question #2714 : Calculus

Example Question #11 : Lines

Example Question #2711 : Calculus

Example Question #242 : Graphing Functions

Example Question #11 : How To Find Equation Of Line By Graphing Functions

Example Question #17 : Equation Of Line

Example Question #1 : Slope

All Calculus 1 Resources

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