Question

The linear approximation for the function f(x)=x^5 near x=2 is 32+80(x-2) Check this.

A.) What number c will give the best quadratic approximation x^5 approximate to 32+80(x-2)+c(x-2)^2. With the help of myininaya I got c=80.

B.)If this approximation is used for various x's in the interval [2,2.1], can you be certian that the error is no bigger than .05? Explain using Taylor's inequality (the Error Bound).

C.) Graph part equation Part A now that we have c. On interval [2,2.1]. note: I don't need someone to graph this, I just want to make sure I have right approach. Would i just pick values like 2, 2.05, and 2.1 for x, set equation equal to y and solve to get points?

Answer 1

f(x)=x^5
f'(x)=5x^4
f''(x)=___?__
f(2)=2^5=32
f'(2)=5(2)^4=5(16)=80
f''(2)=___?__ f(x) approx=f(a)+f'(a)(x-a)
f(x) approx=f(a)+f'(a)(x-a)+f''(a)/2 * (x-a)^2 What is f''(a)/2?

Answer 2

f"(x)= 20x^3
f"(2)=160
f"(a)=80
I'm not sure about the last part after f"(2)

Answer 3

That is right. Now f''(a)/2=?

Answer 4

wait f''(a)=160 not 80?

Answer 5

You mean f''(a)/2=80 correct?

Answer 6

yes