Question

please answer each part with steps included!

3. (10 points) Consider the function f(t) = 32 - 10, and notice that its positive zero is == V10. In this problem, you will use Calculus to estimate 10 to several decimal places. (A) (2 points) Since 3=V9 is close to V10, it is a good place to start. Write down the tangent line to y=f(x) at a = 3. (b) (2 points) Now find the intercept of the tangent line to obtain an estimate for the z-intercept of f(x) and hence of 10. (c) (2 points) Since your answer to part (b) is a better approximation of 10 it is a better place to start. Write down the tangent line to y = f(x) at a = (answer from part (b)) and find its r-intercept. Then repeat the process one more time. You should use a calculator for this part. (d) (4 points) In several complete sentences, explain why you think repeating this process using Linear Approximation produces better and better approximations to the zero.

Answer 1

3. a) f(x) = x2-10: f(3) = 9-10=-). t'(x) = 2x !i! t' (3)= 6* Eqn of Tangent: 7+156 (x-3) 19 = Gre – 19.1 b) x-intercept =)y=0. = 1 . =) Jo 1197 = 3.166667 7 - 3.166667 c) f'(49) = y-36 .. it 12) = 192-10 m (x-1); = 36

19 - 1 - X-inter Cept =) L = 1721 228 3.16 223070 0 3 | at 53; f( ) = 11 44 F( ) = 4 Eqn of tangen: | 9 - 2 - 163963) 114 S1984 Y - Top} =) N = 103963) | 328776 3. 622776601 7 Actual value of Vio = 3.1622776017

d) In this case curve is increasing near the zero,therefore tangent line at point near the zero is below the curve and will intercept x-axis between the the point and zero.

Following successive iteration we again get a point closer to all previous points and zero.