`f(x) = x^3 + 2, 3x - y - 4 = 0` Find an equation of the line that is tangent to the graph of f and parallel to the given line.

Textbook Question

Chapter 2, 2.1 - Problem 36 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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The slope of the line 3x-y-4=0 and the lines parallel to it will be same.

3x-y-4=0

y+4=3x

y=3x-4

Therefore the slope(m) of the line is the coefficient of the x = 3

The slope of the tangent line to the graph f(x) is the derivative of the f(x)

f'(x)=3x^2

Since the tangent line to the graph is parallel to the given line

Therefore 3x^2=3

x^2=1

x=1,-1

The y coordinate can be found by substituting the value of x in the function.

for x=1 y=1^3+2=3

for x=-1 y=(-1)^3+2=1

The equation of the tangent line can be written using the point slope form of the equation.

y-y_1=m(x-x_1)

y-3=3(x-1)

y-3=3x-3 

y=3x

y-1=3(x-(-1)

y-1=3x+3

y=3x+4

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dh246 | (Level 1) Adjunct Educator

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Since the tangent line to function f(x) = x ^3 + 2 has to be parallel to the line 3x - y - 4 = 0, both needs to have the same slope.
Equation of the straight line can be rewritten in slope-intercept form to : y = 3x - 4. Hence the desired slope of our line is 3.

And derivative of f(x) will provide the slope of the line tangent to f(x)
d/dx(f(x)) = 3x
setting it equal to 3
3x = 3 or x = 1 provides one of the conditions

Corresponding y coordinate can be found from the f(x) = x^3 + 2 equation, y = (1)^3 + 2 = 3
The slope of the desired line is equal to 3, and we now have a point on the desired line (1, 3), as well the point of tangency

Using the point-slope, equation to desired line can be obtained as follows. Given slope m = 3 and a point on the desired line,(x0, y0), the equation of the line is given by
(y - y0) = m(x - x0)
y - 3 = 3(x - 1 )
y = 3x

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