Question

If f(1) = 5 and f '(x) 6/x^2+2 , which of the following is the best approximation for f(1.03) using local linearization?

Answer 1

is this what f ' (x) is?

\[\Large f \ ' (x) = \frac{6}{x^2+2}\]

right? or no?

Answer 2

yeah

Answer 3

ok what is f ' (1) equal to?

Answer 4

2?

Answer 5

yes, so m = 2 is the slope of the tangent line

Answer 6

f(1) = 5 means we know (1,5) is on f(x)

so x = 1 and y = 5


y = mx+b
y =2x + b ... plug in m = 2
5 = 2*1 + b ... plug in (x,y) = (1,5)
...
...
...
b = ???

Answer 7

so 3?

Answer 8

yep

Answer 9

L(x) = 2x + 3 is the linear approximating line at (1,5)

so to approximate f(1.03) you need to calculate L(1.03)

Answer 10

i got 5.05

Answer 11

2*1.03+3 = 5.06

Answer 12

thak you so much !

Answer 13

you're welcome