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.