The line y=2x+b is tangent to the graph of y=x^1/2 at the point P=(a, a^1/2). Find P and determine b.

I don't know what to do because they don't give you the x coordinate or anything to plug into the derivative. Help?

y=sqrt x

y'=1/(2sqrtx)

but y=2x+b, slope is 2, so
2=1/(2sqrtx)

sqrt x=1/4

x=+ 1/2

solve for y.

thank you soo much!

To find the point of tangency (P) and determine the value of b, we can use the conditions that the line y=2x+b is tangent to the graph of y=x^1/2.

Step 1: Find the derivative of the function y=x^1/2. This will give us the slope of the tangent line at any given point on the graph.

Let's write the function as y = x^(1/2).

Using the power rule for differentiation, we differentiate y with respect to x:

dy/dx = (1/2) * x^(-1/2).

Simplifying further, we get:
dy/dx = 1 / (2 * sqrt(x)).

Step 2: The derivative of the function represents the slope of the tangent line at any given point on the graph. Since the line y=2x+b is tangent to the graph at point P=(a, a^1/2), the slope of the tangent line should be equal to 2 at this point.

So we have:
2 = 1 / (2 * sqrt(a)).

Simplifying the equation, we get:
4 * sqrt(a) = 1.

Step 3: Solve for a.

To eliminate the square root, we can square both sides of the equation:

(4 * sqrt(a))^2 = 1^2.

Simplifying further:
16 * a = 1.

Dividing both sides by 16:
a = 1/16.

Step 4: Substitute the value of a back into the equation for the line y=2x+b.

Since P=(a, a^1/2), P=(1/16, (1/16)^(1/2)).

Taking the square root:
P=(1/16, 1/4).

Step 5: Determine the value of b.

We know that the equation of the tangent line at P should be y=2x+b. Substituting the x-coordinate for P (1/16) and the y-coordinate for P (1/4), we can solve for b:

1/4 = 2 * (1/16) + b.

Simplifying the equation:
1/4 = 1/8 + b.

Subtracting 1/8 from both sides:
b = 1/4 - 1/8.

Simplifying further:
b = 2/8 - 1/8.

b = 1/8.

Therefore, the point of tangency is P=(1/16, 1/4), and the value of b is 1/8.