Caluclus for bisness help?

Question

I need help solving these problems and if know them plz show or explain how...

1) show that the derivative of x^3 is 3x^2 by using the limit defination of the derivative lim f(x+h)-f(x) all over h. h->0


2) use the method of bi-section to find the root of the equation x^3-x+1=0 accurate to two decimal places.

Hopefully u understand what i wrote!

Answer 1

1.
f(x) = x^3
f'(x) = lim{[(x + h)^3 - x^3]/h} as h-->0
f'(x) = lim{[x^3 + 3x^2h + 3xh^2 + h^3 - x^3]/h} as h-->0
f'(x) = lim{[3x^2h + 3xh^2 + h^3]/h} as h-->0
f'(x) = lim(3x^2 + 3xh + h^2) as h-->0
f'(x) = 3x^2

2.
x^3 - x + 1 = 0
Finding the x at which the function first goes negative,
- 1 + 1 + 1 = 1
- 8 + 2 + 1 = - 5
so there is a root between -1 and -2
(-1 - 2)/2 = -1.5
f(-1.5) = -0.875
(-1.5 - 1)/2 = -1.25
f(-1.25) = 0.296875
(- 1.5 - 1.25)/2 = -1.375
f( -1.375 ) = -0.224609375
( -1.375 - 1.25)/2 = -1.3125
f( -1.3125 ) = 0.114013671875
( -1.375 -1.3125 )/2 = -1.34375
f( -1.34375 ) = -0.082611083984375
( -1.34375 -1.3125 )/2 = -1.328125
f( -1.328125 ) = -0.014575958251953125
( -1.328125 -1.3125 )/2 = -1.3203125
f( -1.3203125 ) = 0.018710613250732421875
( -1.328125 -1.3203125 )/2 = -1.32421875
f( -1.32421875 ) = 0.002127945423126220703125
so, to 2 decimal places, we have
x ≈ -1.32
(All done with the Windows calculator. Doing it by decimal places is easier & faster using a spreadsheet! This many operations would produce an answer to 8 decimal places.)

Answer 2

I will help you with the first one..
You first want to set up your limit:

lim[h->0] ((x+h)^3 - x^3)/h ; since division by zero is a no no, we must simplify the equation so that we are not dividing by zero:

expanding (x+h)^3 makes the equation:
(x^3 + 3x^2*h + 3x*h^2 + h^3 - x^3 )/h

subtracting x^3 and factoring out all of the h's gives us

= 3x^2 + 3xh + h^2

Great, now we don't have h as a denominator anywhere in the new function so we can take the limit:

lim[h->0] 3x^2 + 3xh +h^2 = 3x^2 + 3x(0) + (0)^2 = 3x^2

As far as #2, I am not completely familiar with the method of bi-section.. maybe you could explain it better?

Answer 3

I can do 1 here, but 2 needs to be programmed.

Derivative of x^3 + 3x^2 =

Lim(h->0){[(x+h)^3 + 3(x+h)2 - x^3 - x^2]/h} =
(by multiplying out and simplifying)

Lim(h->0){[3hx^2 + 3xh^2 + h^3 + 6hx + 3h^2]/h} =

Lim(h->0){[3x^2 + 3xh + h^2 + 6x + 3h} =

3x^2 + 6x

Answer 4

i know the answer, I just cant explain it here.
You can use the power rule and the constant multiple rule too if you know them. But you have to do this the long way for your teacher.