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What is the equation of the line tangent to #f(x)= sqrt(3-2x) # at #x=-2#?

Answer 1

Find the first derivative to get the slope of the tangent line. Then use the point slope form to get the general equation of the tangent line.

First, find the first derivative because this will give you the slope of the tangent line.

#dy/dx= 1/2(3-2x)^(-1/2) (-2)#

Or simply:

#dy/dx=-(3-2x)^(-1/2)#

Substituting #x#:

#dy/dx = -(3+4)^(-1/2)#

#dy/dx = - (-7)^(-1/2)#

#dy/dx = (7)^(-1/2) = 1/7^(1/2) = 1/sqrt7##

#dy/dx = sqrt(7)/7#

The slope of the tangent line is #sqrt(7)/7#.

You have #x= -2#.

Solving for #y#:

#y = sqrt(3-2(-2))# #y = sqrt (3+4)# #y = sqrt (7)#

Using the point-slope form:

#y-sqrt(7)= sqrt(7)/7(x-(-2))# #y-sqrt(7) = sqrt (7)/7(x+2)#

You can put it in slope-intercept form by solving for #y#:

#y = sqrt7/7x+(9sqrt7)/7#

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Answer 2

Axel Alvarez

The equation of the line tangent to f(x) = sqrt(3-2x) at x = -2 is y = -2x + 7.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.