+0  
 
0
67
1
avatar+16 

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
Sort: 

1+0 Answers

 #1
avatar+86626 
+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 )

 

 

 

cool cool cool

CPhill  Mar 13, 2018

28 Online Users

avatar
avatar
New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy