Calculus question?

Question

(a) If G(x) = x/(1 + 2x), find G '(a).

and

Use it to find an equation of the tangent line to the curve y = x/(1 + 2x) at the point (-1/4, -1/2).

Answer 1

This is just using the quotient rule, which is "if f(x) = g(x)/h(x), then f'(x) = (h(x) * g'(x)) - (g(x) * h'(x)) divided by h(x) squared". So in this problem, you get this as the derivative:

((1+2x)(1) - (x)(2)) / (1+2x)^2

which is simplified as 1 / (1+4x + 4x^2)

Now, to find the tangent, you plug the given points into your derivative.

G'(-1/4) = 1/(1 - 1/2)^2
G'(-1/4) = 4
Slope = 4

Next you set up the equation for the line in point - slope form, or (y-y1) = (Slope)(x-x1)

y + 1/2 = 4(x + 1/4)
y + 1/2 = 4x + 1
So your final equation for the line equals y = 4x + 1/2

Answer 2

G(x) = x/(1 + 2x)
G'(x) = 1/(1+2x) -2x/(1+2x)^2 = 1/(1+2x)^2
so
G'(a) = 1/(1+2a)^2

We need to find the slope at a = -1/4.
G'(-1/4) = 1/(1 - 1/2)^2 = 1/(1/4) = 4

Let m be the slope. Then the tangent will have an equation of the form y = mx + b = 4x +b. We know that (-1/4, -1/2) is on the line so

-1/2 = 4(-1/4) + b
b = -1/2 +1 = 1/2

y = 4x + 1/2 is the equation of the tangent

Answer 3

G'(x) = [(1+2x)-2x]/(1+2x)^2 = 1/(1+2x)^2
G'(a) = 1/(1+2a)^2
G'(-1/4) = 1/(1+2(-1/4))^2 = 1/(1/2)^2 = 1/(14) = 4
So the tangnet has the form y=4x +b where 4 is the slope at x = -1/4 and b is the t -intercept
When x = -1/4, y = (-1/4)/(1+2(-1/4)) = -1/2
so -1/2 = 4(-1/4)+b --> b = 1/2
So equation is y = 4x + 1/2

Answer 4

omg that looks nearly impossible.

Good luck with that one.