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

StayCurly
Mar 13, 2018

#1**+2 **

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 )

CPhill
Mar 13, 2018