Find the linear approximation to f(x)= (1)/(x^2 + 1) at x=1and use it to approximate f(1.1)

Question

anonymous · Answer

The idea behind linear approximation is that you want to draw a line that is really close to the curve at a specific point so you can estimate the point on the curve. Since they told us to find the linear approximate of f(1.1), then the tangent line at f(1) will make a great starting point. 

remember y=mx + b is the equation for a line

we have an (x) value, and we can plug it in to find the (y) value:

f(1) = 1/2

so you're point will be (1, 0.5)

Remember that the definition of a derivative is just the slope of the tangent line? well then all we have to do is find the derivative and we can write our linear equation out.

$$f(x) = 1/(x^2 + 1) $$
$$f'(x) = -2x/(x^2 + 1)^2$$

lets plug in the x=1 to find the slope of our line

$$f'(1) = -2/4 = -1/2 = -0.5$$

so our equation is:

y= -0.5 (x) + b

to find b, we can plug in our ( 1, 0.5)

0.5 = -0.5 + b

b = 1

so our linear equation is:

y = -0.5 (x) + 1

now we can just plug in our x=1.1 to find f(1.1)

y = -0.5 (1.1) + 1
y = -0.55 + 1
y = 0.45

so the linear approximate of f(1.1) is about 0.45