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!
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
To find the values of x where the tangent line to the graph of the equation is horizontal, we need to determine where the derivative of the equation is equal to zero.
First, let's find the derivative of the equation y = x - sqrt(x).
Differentiating both sides with respect to x, we get:
dy/dx = d/dx(x) - d/dx(sqrt(x))
dy/dx = 1 - (1/2)(x)^(-1/2)
Now, let's set the derivative equal to zero and solve for x:
0 = 1 - (1/2)(x)^(-1/2)
To simplify this equation, let's multiply both sides by 2:
0 = 2 - (x)^(-1/2)
Now, add (x)^(-1/2) to both sides:
(x)^(-1/2) = 2
To eliminate the square root, we can square both sides of the equation:
[(x)^(-1/2)]^2 = 2^2
x^(-1) = 4
1/x = 4
Now, we can solve this equation for x by taking the reciprocal of both sides:
x = 1/4
Therefore, the only value of x where the tangent line to the graph of the equation y = x - sqrt(x) is horizontal is x = 1/4.