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# Derivative question

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Let $$f(x) = 4\sqrt(x) - x$$

a) Find all points on the graph of f at which the tangent line is horizontal
b) Find all points on the graph of f at which the tangent line has slope -1/2

I kinda understood how the first one worked, but the second question threw me for a loop. How would I work through both?
Thanks guys

Mar 13, 2018

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f(x)  = 4x^(1/2)   - x

f ' (x)   =  (1/2*4x^(-1/2)  - 1  =    2x^(-1/2)  - 1

The tangent line will be horizontal if  the slope  = 0

So.....set the derivative equal to 0 and solve for x

2x^(-1/2)  -  1  = 0

2x^(-1/2)  = 1

x^(-1/2)  = 1/2

We can write this as

x^(1/2)  = 2        square both sides

And  x  =  4

When  x  =  4 ,   y  = 4√4  - 4   =  8 - 4  =  4

So.....the point where the tangent line is horizontal is  (4,4)

To find where the slope is  -1/2....we have

2x^(-1/2)  -  1  = -1/2

2x^(-1/2)  =  1/2

x^(-1/2)  =  1/4

And we can write this as

x^(1/2)  =  4     square both sides

x  = 16

When  x  = 16 ,   y  = 4√16  - 16   =  0

So....the point where the tangent line  = -1/2  = (16, 0 )

Mar 13, 2018