calculate the slope of tangent to the given function at the given point
f(x) = 2 / √x+5 , P(4,2/3)
I assume you meant:
f(x) = 2/√(x+5) = 2(x+5)^(-1/2)
f ' (x) = -(x+5)^(-3/2)
at (4, 23)
f ' (4) = -(9)^(-3/2)
= -1/9^(3/2) = -1/27
tangent equation:
(y-2/3) = (-1/27)(x-4)
27y - 18 = -x + 4
x + 27y = 22
To calculate the slope of the tangent to the function at a given point, we need to find the derivative of the function and then substitute the x-coordinate of the given point into the derivative.
1. Find the derivative of the function f(x):
To calculate the derivative of f(x), we will apply the power rule and the chain rule.
f(x) = 2 / √x+5
Using the power rule, we can rewrite the expression as:
f(x) = 2(x+5)^(-1/2)
Applying the chain rule, we differentiate the expression:
f'(x) = [2 * (-1/2) * (x+5)^(-1/2-1)] * (1)
Simplifying the expression:
f'(x) = -1 / √(x+5)^3
2. Substitute the x-coordinate of the given point P(4, 2/3) into the derivative.
f'(4) = -1 / √(4+5)^3
= -1 / √(9)^3
= -1 / 27
Therefore, the slope of the tangent to the function f(x) = 2 / √x+5 at the point P(4, 2/3) is -1/27.