# let f(x) = (x-1)^2, g(x) = e^-2x and h(x) = 1 +ln(1-2x). (a). Find the linearizations of f, g and h at a=0. What do you notice? How do you explain what happened? (b). Graph f, g, and h and their...

let f(x) = (x-1)^2, g(x) = e^-2x and h(x) = 1 +ln(1-2x).

(a). Find the linearizations of f, g and h at a=0. What do you notice? How do you explain what happened?

(b). Graph f, g, and h and their linear approximation. For which function is the linear approximation best? For which is it worst? Explain.

The linear approximation is -2x +1 for all functions. What is happening?, and also question (b) I am confused on. It would be a straight line graphically slope of -2 and crossing through points (1,1)???? not sure how to graph f(x), g(x), h(x)????

### 1 Answer | Add Yours

As you found, the linear approximation for all three is the same; y=-2x+1. This means that the instantaneous rate of change of all three functions at x=0 is the same.

(b) Graph `y=(x-1)^2` and the linear approximation y=-2x+1; the graph of f(x) is a parabola opening up with vertex (1,0):

Graph `y=e^(-2x)` and the linear approximation; the graph of g(x) is an exponential decay:

Graph `y=1+ln(1-2x)` and the linear approximation; the graph of h(x) is teh graph of the natural logarithm shifted 1 unit up, one unit left, reflected over the vertical axis and vertically stretched by a factor of 2:

If you use some sort of graphing utility and "zoom" in around x=0, you will see that the line is a really good approximation for the quadratic as long as -1/2<x<1/2. (or you can plot points on graph paper)

To verify algebraically that the approximation is best for the quadratic, you can test a point, say x=1/4 and note the difference from the actual value and the approximate value.

`f(1/4)=(1/4-1)^2=9/16` compared to `-2(1/4)+1=.5` for a difference of 0.0625.

`g(1/4)~~.607` compared to .5 for a difference of .107

`h(1/4)~~.306` compared to .5 for a difference of .194

The smallest difference is between the quadratic and the linear approximation.