Let G(x)=2/(x+1)
(a) Use the limit definition f the derivitive to find G'(0).
(b) Find an equation of the tangent line to the curve y=2/(x+1) at the point (0,2).
I will be happy to critique your work. There is a bit of algebra involved with this.
Just not really sure where to start with this problem
Answer
To find the derivative of a function using the limit definition, we can follow these steps:
(a) Use the limit definition to find G'(0):
Step 1: Start with the limit definition of the derivative:
G'(0) = lim┬(h→0)〖(G(0+h)-G(0))/h〗
Step 2: Substitute the expression for G(x):
G'(0) = lim┬(h→0)〖(2/(0+h+1)-2/(0+1))/h〗
Step 3: Simplify the expression:
G'(0) = lim┬(h→0)〖(2/(h+1)-2/1)/h〗
G'(0) = lim┬(h→0)〖(2/(h+1)-2)/h〗
Step 4: Find a common denominator:
G'(0) = lim┬(h→0)〖(2-2(h+1))/(h(h+1))〗
G'(0) = lim┬(h→0)〖(-2h)/(h(h+1))〗
Step 5: Simplify the expression:
G'(0) = lim┬(h→0)〖-2/(h+1)〗
Step 6: Evaluate the limit as h approaches 0:
G'(0) = -2/(0+1)
G'(0) = -2/1
G'(0) = -2
Therefore, G'(0) is equal to -2.
(b) To find the equation of the tangent line to the curve y = 2/(x+1) at the point (0,2), we can use the slope-intercept form of a linear equation, y = mx + b, where m is the slope and b is the y-intercept of the tangent line.
Step 1: Find the derivative of G(x):
G(x) = 2/(x+1)
G'(x) = -2/(x+1)^2
Step 2: Plug in the x-coordinate of the point (0, 2) into G'(x) to find the slope:
m = G'(0) = -2
Step 3: Plug the slope and the coordinates of the point (0, 2) into the slope-intercept form of the equation:
y = mx + b
2 = -2(0) + b
2 = b
Therefore, the equation of the tangent line to the curve y = 2/(x+1) at the point (0,2) is y = -2x + 2.