+31 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
+128475 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 )
