math help!!?

Question

I have a basic question on calculus ....please can someone explain to me :


Let f(x) = x^2

y + delta y = f ( x + delta x )

The book says that a small increase in x will cause a small increase in y.

But if I put lets say 3 and 2 for x and dx y is 5^2 which is 25.

But this isn't the same as f(3) + f(2) which equals 13 ?

What have I done wrong ?




Also, can f ( a+b+c ) be treated as f(a) + f(b) + f(c) ?


Thanks a lot for any help

Answer 1

if you mean y = x^2
then differenciating y with respect to x,
dy/dx = 2x
let x = 2
therefore, dy/dx = 2 * 2 = 4
and y = 4
now let delta x = 0.001 (yes, they are talking about that kind of small change, not jumping directly from numbers like 3 to 2 because we are dealing with small numbers here - it ofcourse depends upon how greatly y varies with x - if the variation is very, very small, then ofcourse dy/dx is very small and in that case the change in x could be as large as 2 to 3 - so what you are doing wrong here is taking an extremely big change in x or delta x)

For small change, we have,
dy/dx approximately equal to delta y / delta x
therefore, 4 = delta y / 0.001 (approximately)
or, delta y = 0.004
therefore new y = 4 + 0.004 = 4.004
so, the book is right, a small change in x did cause a very small change in y.
And look at this carefully, delta x here is 0.001
so,
f (2 + 0.001)= f (2.001) = 4.004001, as seen in the computer calculator, so my calculation using calculus is correct - as is the books statement = y + delta y,
which is = 4 + delta y
and according to my calculation as shown above, delta y = 0.004
therefore y + delta y = 4.004,
which is equal to f(x + delta x)
and no, f(a+ b+ c) is not equal to f(a) + f(b) + f (c)

Answer 2

Part 1:

Not sure if you have provided all info. But delta x is supposed to be very very small compared to x. So you cannot have x = 3 and delta = 2.

Say y = f(x) = x^2.
when x = 3, y = f(x) = 9.
Take delta x = 0.01
So f (x + delta x) = f(3.001) = (3.001)^2 = 9.006001

Now y + delta y = f (x + delta x) = 9.006001
So delta y = 9.006001 - y = 9.006001 - 9
= 0.006001

So the saying of small increase in y is correct.

Part 2:

No. f (a + b + c) is not the same as f(a) + f(b) + f(c). You just need to think f (x) = x^2 and you know it is not the same.

Eg. a = 1, b = 2, c = 3.

f (a+b+c) = f (6) = 36
f (a) + f(b) + f(c) = 1 +4 +9 = 14.

Answer 3

First of all, small increalse means REALLY SMALL INCREASE, which 2 is not - it is very much signficant, and delta x should be close to zero.

Example: f(x)=x^2
x=3
f(x)=9
Delta x =0.001
f(3.001)=9.006001 - isn't thisincrease small enough to prove the postulate You have menitioned?

As for f(a+b+c) - it is not always equal to f(a)+f(b)+f(c)
example: let f(x)=x^2
Let a=1, b=2, c=3
f(1+2+3)=f(6)=36
f(1)+f(2)+f(3) = 1+4+9 = 14

Your statement works for short linear function, i.e. functions that involve only one linear variable, without a free member (linear function is y=ax+b, where b is the free member). For all other functions it is incorrect.

Answer 4

since f(x+y) not equal 2 f(x)+ f(y)


considering a function with a power 2

ie f(x+y)= x^2 + y^2 + 2xy =/= x^2 + y^2

this is a simple solution but u need practise 2 understand this

sam ;)

Answer 5

write out f(x + dx) = x^2 + 2x*dx + dx^2
y + dy = x^2 + 2x*dx + dx^2
so your dy = 2xdx+dx^2

take dx is very small say dx = x/100

Answer 6

Typically f(a+b+c ) is not equal to f(a) + f(b) + f(c)
some exampes when it is true is
f(x) = 2*x
f(x) = x/5
But in general this is not true, for e.g.,
f(x) = x^2
f(x) = x+2
f(x) = 9x^2+3x+4
etc

Answer 7

f(a + b + c) is not equal to f(a) + f(b) + f(c)

I will just give an example
if f(x) = sin x
f(a + b + c) = sin (a + b + c)
f(a) + f(b) + f(c) = sin a + sin b + sin c

You see the difference?

Answer 8

girlfriend, i wish i could help you, i am out in the cold on this but i will ask my husband when comes home. good luck!!!

Answer 9

No! Because f is not linear!
f(x+y) =/= f(x) + f(y)
because
f(x+y)=(x+y)²=x²+2xy+y²
And if 2xy =/= 0 then f(x+y) =/= f(x) + f(y) !