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# Derivatives from tangent lines

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Suppose the line tangent to the graph of f at x = 2 is y = 4x + 1 and suppose the line tangent to the graph of g at x = 2 has slope 3 and passes through (0, -2). Find an equation of the line tangent to the following curves at x = 2.
a. y = f(x) + g(x)

b. y = f(x) - 2g(x)

c. y = 4f(x)

I don't even know where to start with this one. Thanks for helping!

Mar 13, 2018

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Suppose the line tangent to the graph of f at x = 2 is y = 4x + 1

This tells you that the point (2,9) is on the f graph

$$f(2)=9$$

The Equation of tangent to the f funtion    at x=2   is    y=4x+1

and suppose the line tangent to the graph of g at x = 2 has slope 3 and passes through (0, -2).

At   f'(2) = 3

And this tangent line, the one at f(2)  passes through (0,-2)

For  this line we are told (0,-2) is a point and the gradient is 3.    We need to find the find the y value for x=2.

$$m=\frac{rise}{run}\\ 3=\frac{y_1--2}{2-0}\\ 6=y_1+2\\ y_1=4\\ g(2)=4$$

The Equation of tangent to the g funtion    at x=2   is    y=3x-2

a. y = f(x) + g(x)

f'(2)+g'(2)=4+3=7

f(2)+g(2)= 9+4=13

So the equation of the tangent at x=2 is

y=7x+13

b. y = f(x) - 2g(x)

2g(x) the gradient and the y values will be doubled    so gradient of tangent=6 at (2,8)

f'(2)+2g'(2)=4+6=10

f(2)+g(2)= 9+8 = 17

y=10x-17

c. y = 4f(x)

tangent of 4f(x)=4(4x+1) at x=2     is   y = 16x+4

I think that is right.

You can try convincing yourself (one way or the other) by drawing some graphs.

Mar 14, 2018