calculus quest?

Question

let f(x)=Ax + (B/x^2) where A and B are constants. Find values for A and B so that the line y=2x+1 is tangent to y=f(x) when x=-1.

I have no idea how to do this./

i keep getting different answers and different ideas how to do this...none of the answers work out to be tangent and the problem is really getting annoying...please help!

whatever i get the graph is not a tangent

I understand the logic...get two freaking equations and solve them ... i keep getting things like A=4/3 or A=1/3 etc but no matter what, it is not a tangent when graphed...whats going wrong?

Answer 1

the slope of the tangent to f(x) is = df(x)/dx
here, differentiating f(x) with respect to x we get
df(x)/dx = x -2B/x^3
for x = -1 the value of df(x)/dx is = 2B - 1
we know that the slope of the line given is 2
so, 2B - 1 = 2
or, B = 3/2
and at x = -1 the y coordinate of both the tangent (line) and f(x) must be equal, so f(-1) = -A + 3/2 must be equal to 2(-1) + 1
so, A = 5/2

Answer 2

You're given the tangent line to a curve at a given point. That actually gives you 2 pieces of information.

#1. The slope of the line is the derivative of the curve at that point.
#2. Both the line and the curve contain the given point.

So you have

f'(-1) = slope of line
f(-1) = value of line at x = -1

You have 2 (linear) equations and 2 unknowns to help you solve the problem.

Answer 3

f(x) = b*x^-2 + a*x

f'(x) = -2b*x^-3 + a, which gives you the slope at a point.

f'(-1) = -2b*(-1)^3 + a = 2b + a = 2.

f(-1) = b - a and when x = -1, 2(-1) + 1 = -1 --> b - a = -1.

Given 2b + a = 2 and b - a = -1 you can find the values of a and b.