You have the right idea, but
y' = 1 - 1/(2√x)
Setting that to zero, we get
1 - 1/(2√x) = 0
1/(2√x) = 1
2√x = 1
√x = 1/2
x = 1/4
As you can see from the graph, there is only the one point (1/4,-1/4) with a horizontal tangent.
http://www.wolframalpha.com/input/?i=x-%E2%88%9Ax
Find all values of x if any where the tangent line to the graph of the given equation is horizontal?
y= x - sqrt (x)
I am very confused would I work it like this?
y = x - x^-1/2 then set x = 0?
Please explain. thank you!
1 answer