Q: Find the linearization of f(x) = ex at x = 0
A: First, some concept: Linearization is the act of finding a “linear function” that can approximate the given function on or around a given point. In this problem, we want to find a line that models the shape of ex when you are around the point x = 0.
Step 1: Find an (x, y) point on the function in question.
The function is f(x) = ex . The x part of the point is 0. Plug that in to find y:
f(0) = e0 = 1
So, the point is (0, 1). Hold this point. We will need it for later.
Essentially, we want to find a line that follows that patterns of f(x) = ex and goes through the point (0, 1).
So, to model the pattern, we need to slope of f(x) = ex at the given point.
Step 2: Find the derivative (slope) of the function at the given point:
f(x) = ex
Find the derivative:
f ‘ (x) = ex
Now, find the slope at (0, 1):
f ‘ (0) = e0 = 1
So, the slope is 1. m = 1
Step 3: Find a line that has the same slope as the function that goes through the given point:
We need a line with slope of 1 that goes through the point (0, 1).
Start with your equation of a line:
y = mx + b
We know m = 1, so:
y = x + b
Plug in the (x, y) point to find “b”
1 = 0 + b
1 = b
So, the equation of the line is y = x + 1
In summation, to approximate values of f(x) = ex around where x = 0 you can use the line y = x + 1