Question

If f(-1) = -3 and f'(x)=(4x^2)/(x^3+3), which of the following is the best approximation for f(-1.1) using local linearization?

Answer 1

-7.2
2.8
-1.2
-3.2

Answer 2

hey, just curious... do u understand the theory of solving ODE's (1st and 2nd)?

Answer 3

not preaching, just if u do, i can help, if not.... it's just after 5:30am here, and i've been awake all night, so imma go to sleep pretty soon

Answer 4

ohh ok I dont but thats ok

Answer 5

ok, im going to assume that's a yes?
1st order linear ODE
exact solution is:
f(x) = 1/3 (4 log(x^3+3)-9-(4 log(2)))
so plug x = -1.1 into this
solve.

u ok with that?

Answer 6

sorry... @zepdrix are you able to assist @chris215 with the step by step, I'm pretty much crashing hard right now

Answer 7

So for linear approximation we use something like this, ya?\[\large\rm f(x)\approx f(x_0)+f'(x)(x-x_o)\]
To approximate our function f at some x,
we calculate a nearby point f(x_o) and then add some stuff to it.

For this problem, our point that we're approximating is at \(\large\rm x=-1.1\).
We're using a place nearby to approximate \(\large\rm x_o=-1\).

So for our approximation, f(-1.1) is going to be about this much,\[\large\rm f(-1.1)\approx f(-1)+f'(-1.1)(-1.1-(-1))\]

Answer 8

We can simplify the last part in the brackets,\[\large\rm f(-1.1)\approx f(-1)-0.1\cdot f'(-1.1)\]

Answer 9

They gave us information to figure the rest out,
f(-1)=-3
and
f'(x)=4x^2/(x^3+3)

So plug in your -3 in place of f(-1),
and also, evaluate f'(x) at x=-1.1, and plug it in.

Answer 10

\[\Large L(x)=f(a)+f'(a)(x-a)\]

\[\Large L(-1.1)=f(-1)+f'(-1)(-1.1-(-1))\]

Answer 11

notice that \(f'(x)\) is evaluated at -1 and not -1.1

Answer 12

woops :))

Answer 13

Thanks guys :))