Hi Alex,
Anytime a calculus book or teacher says to find the linear approximation of a function centered at a=#, that means you want to find the tangent line at the function. So first find the derivative of the function. g'(x)=(1/5)(x+1)-4/5. Now plug in x=0, because you want the line tangent to the graph at x=0. You should get g'(0)=1/5.
Now that is the slope of your line. Next, you need an ordered pair to find the equation of the line. a=0 means the x-value is 0. So find the y-value by using g(0)=1. So the ordered pair is (0,1). Using the point slope form, you get y-1=(1/5)(x-0). Solve for y. You get y=(1/5)*x+1 which is the tangent line you wish to graph. Now to approximate the fifth root of .95 and 1.1, you need to figure out which x-value you are going to plug into the line that you just found. so set (x+1)1/5=.951/5 and solve for x. You should get -.05 which is what you will plug into the line. y=(1/5)*-.05+1=-0.01+1=.99. Now do the same for 1.11/5.
