
John G. answered 11/13/19
Physics, Math, Chemistry teacher 30 years
for this approximation we start at x=2 and then move forward or backward using the slope of the tangent line at 2. so g(1.99) = g(2) + (-0.01)(the derivative evaluated at 2).
Since the derivative at 2 is sqrt(22 +5) we calculate it as 3.
so , now g(1.99) = g(2) + (.01)*3) or -1 + -.03 = -1.03
similarly, g(2.01) = -1 + (3))(.01) or -0.97
For part 2. i am not sure what at a=0 refers to. lets assume we want to approximate from x=0. Then f(0) = 1.
Now, we must find the derivative odf f(x) which is 1/3 (1+x)-2/3 . If we plug in 0 the slope is 1/3.
so, f(-0.05) = 1 + (-0.05)(1/3) = 0.983
f(0,1) = 1 + .1(0.3333) = 1.033

John G.
For .95 I will use x=-.05 so that the 1+x will be .9511/14/19
Rahman W.
Thank you for your help, for part 2 shouldnt you use the linear approx method, that way you will get, -0.33x-1? I just dont know how to approximate the numbers of (0.95)^1/3 and (1.1)^1/3 after I find this.11/14/19