Find the linear approximation of f(x)= (1+x)^1/5 at a=0 and use it to approximate the numbers (0.98)^1/5 and (1.01)^1/5

linear approximation = line with the same slope as f(x) at a=0, which intersects with f(x) at 0.

slope=f'(x)

derive f(x) to get (1/5)(1+x)^(-4/5) so f'(x)=(1/5)(1+x)^(-4/5)

plug in point 0 to get slope at 0 = 1/5

to create an equation for a line you need to use the formula (y1-y2)/(x1-x2)=slope, so you need one point on the line. Luckily, you have the equation for f(x) and so you can find where f(x) is at 0, and since f'(x) intersects (is tangent to f(x)) at 0, you can use the point on f(x) as a point on f'(x).

f(0)=(1+0)^1/5 = 1. so the point is (0,1)

plug it in to the formula, and you get (y-1)/(x-0)=1/5. mulitply out the x and you get y-1=x/5, which is your equation for you line.

Now all you have to do is plug in the different points.