if you use the formula, y= f(a)+ f'(a)(b-a), i'm not getting the right answer, i'm following the example we did in class, but its not coming out right, what am i doing wrong?
Your formula is correct. In this case you have a = 1, b = 1.03 and f(x) = x^12 + x. The derivative is f'(x) = 12x^11 + 1 and so f'(a) = f'(1) = 13. Further f(1) = 2. So the approximate value is y = 2 + 13(1.03 - 1) = 2.39. How does this differ from your calculation?
what values should i make my (a) and (b) for this problem i made (a)=.9 and (b)=.99 and i cannot get the right answer
The problem says to use the tangent line approximation at t = 1. This means that a should be 1. Your choice of b is correct.
One has c(1) = 16 and c'(t) = 8(21.4)t^(20.4) + 3 and so c'(1) = 174.2. The tangent line is y = c(1) + c'(1)(t - 1) = 16 + 174.2(t - 1). Let t = .99 to get y = 16 + 174.2(-0.01) = 16 - 1.742 = 14.258.
How do you do these types of problems?
First calculate the tangent line approximation as in the other problems. Use a = 3.0 and f(x) = x^2 + 4x. The tangent line at (a, f(a)) is given by y = f'(a)(x-a) + f(a). Since f(a) = f(3) = 21 and f'(x) = 2x + 4, one has f'(a) = f'(3) = 10 and the tangent line is y = 10(x-3)+21.
Since you expect that for x near 3, the tangent line will be a good approximation to the graph of the function y = f(x), one can approximate the value of x where f(x) =22, by looking to see where the tangent line has y = 22. So, we solve 10(x-3)+21 = 22. One should get x = 3.1.