Differentiation - Help, Please!

[1calculus1]

Differentiation ---Help, please!

1. Homework Statement
1) Use Newtons Method to approximate the real zero of the function f(x)=x^3 + 7x + 3 = 0

2) The curve y = x^3-2x^2+x-3 intersects the curve y = cos(2x). If x = 1.8 is used as the first estimate then, using Newton's Method, what is the x-value or the intersection point, accurate to 4 decimal places?

3) A curve is given by y = a^3, where a = m^2(2m-1). Approximate the slope of the curve when m = 0.6543

4) If y = -4 / $$\sqrt[3]{x+5}$$, then dy/dx

5) Find an equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1

2. Homework Equations
1) Newtons Method:
x = f(x) / f'(x)
Or Slope = Rise/Run

2) Newtons Method
x = f(x) / f'(x)
Or Slope = Rise/Run

3) y = mx+b
derivatives

4) Quotient Rule
(f'g-g'f) / (g^2)

5) Quotient Rule
(f'g-g'f) / (g^2)
y=mx+b

3. The Attempt at a Solution
1) x^3 + 7x = 0
Let the "guess" be "n"
Let the next "guess" to be "n2" (2 was suppose to be a subscript but I can't find it in the Latex Reference)
Slope: derivative of x^3 + 7x = 0
3x^2 + 7 = 0
Rise: Plug in the "n" to the equation
n^3 + 7n = 0
Run: "n" subtract the "n2"
n - n2
Final formula:
3x^2 + 7 = n^3 + 7n / n - n2
*The answer should be this: (I was close but my rise is really off)
The first "n" is suppose to be a lower subscript. I can't find it here in the Latex Reference.
xn +1 = 2x$$^{3}_{n}$$ -3 / 3x$$^{2}_{n}$$ + 7

2) x^3 - 2x^2 + x - 3 - cos(2x) = 0
Slope = Rise/Run
Slope: Derivative of x^3 - 2x^2 + x - 3 - cos(2x)
3(1.8)^2 - 4(1.8) + 1 + sin2(1.8)(2)
Rise: Plug in 1.8 to the equation
(1.8)^3-2(1.8)^2 + 1.8 -3 - cos(2(1.8))
Run: 1.8 - x
Final Equation:
3(1.8)^2 - 4(1.8) + 1 + sin2(1.8)(2) = (1.8)^3-2(1.8)^2 + 1.8 -3 - cos(2(1.8)) / 1.8 - x
Simplify: 3.645639188 = -2.846026728 / 1.8 - x
-2.846026728 / 3.645639188 = 1.8 - x
1.8 + 0.780666045 = x
x = 2.680666046
The answer in the book is 2.0923.. So, do you think I should use my "x" to be the second trial and do the whole equation again from the start to get the 2.0923?

3)
a = m^2(2m-1)
(0.6543)^2(2(0.6543)-1))
a= 0.13211428
y = a^3
Therefore...
y = 0.13211428^3
y =0.002305946

I'm stuck. Please help?

4) ((0)($$\sqrt[3]{x+5}$$)) - ((-4)(1/3(x+5)^-2/3)) / ($$\sqrt[3]{x+5}$$)^2)
Simplify: ((4)(1/3(x+5)^-2/3) / ($$\sqrt[3]{x+5}$$)
Then I don't know how to simplify it much longer.. My simplifying skills has gone bad, I guess. Help, please?

5) Get y-value, plug in the x-value which is 1
1-1 / 1+1 = 0
(1,0)
Do quotient rule on the equation...
(1)(x+1) - (x-1)(1) / (x+1)^2
Simplify...
(x+1) - (x-1) / (x^2 + 2x+2)
Plug in the x-value which is 1 to get the slope.
(1+1) - (1-1) / (1^2 +2(1) + 2)
2 / 1+2+2 = 2 / 5
Slope = 2/5

y=mx+b
0 = 2/5(1) + b
0 - 2/5 = b
-2/5 = b
Final equation...
y = 2/5x -2/5

Although it seems like I did everything right, my final equation seems to be incorrect from the answer in the book which is x-2y=1





THAT'S ALL FOR NOW. I'll continue more questions later on if I find any that bugs my mind. But for now, I need to go to church so in the meanwhile, PLEASE HELP? =/

 

[1calculus1]

Anyone there?

 

[1calculus1]

Please?

 

[1calculus1]

At least just help me with question #3.. PLEASE?

 

[HallsofIvy]

1. Homework Statement
1) Use Newtons Method to approximate the real zero of the function f(x)=x^3 + 7x + 3 = 0

2) The curve y = x^3-2x^2+x-3 intersects the curve y = cos(2x). If x = 1.8 is used as the first estimate then, using Newton's Method, what is the x-value or the intersection point, accurate to 4 decimal places?

3) A curve is given by y = a^3, where a = m^2(2m-1). Approximate the slope of the curve when m = 0.6543

4) If y = -4 / $$\sqrt[3]{x+5}$$, then dy/dx

5) Find an equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1

2. Homework Equations
1) Newtons Method:
x = f(x) / f'(x)
Or Slope = Rise/Run

That's the definition of "slope", not "Newtons method". Newton's method says $x_{n+1}= x_n- f(x_n)/f '(x_n)$

[/itex]2) Newtons Method
x = f(x) / f'(x)
Or Slope = Rise/Run

3) y = mx+b
derivatives

4) Quotient Rule
(f'g-g'f) / (g^2)

5) Quotient Rule
(f'g-g'f) / (g^2)
y=mx+b

3. The Attempt at a Solution
1) x^3 + 7x = 0
Let the "guess" be "n"
Let the next "guess" to be "n2" (2 was suppose to be a subscript but I can't find it in the Latex Reference)
Slope: derivative of x^3 + 7x = 0
3x^2 + 7 = 0
Rise: Plug in the "n" to the equation
n^3 + 7n = 0
Run: "n" subtract the "n2"
n - n2
Final formula:
3x^2 + 7 = n^3 + 7n / n - n2
*The answer should be this: (I was close but my rise is really off)
The first "n" is suppose to be a lower subscript. I can't find it here in the Latex Reference.
xn +1 = 2x$$^{3}_{n}$$ -3 / 3x$$^{2}_{n}$$ + 7

You understand, don't you, that a "real zero of the function f(x)=x^3 + 7x + 3 = 0" is a number? Applied to this problem, f(x)= x³+ 7x+ 3 so the derivative is 3x²+ 7. Newton's method says that if x_n is a close to a solution then $x_{n+1}= x_n- (x^3+ 7x+ 3)/(3x^3+ 7)$ will be closer. If you take x₀= 0, then $x_1= 0- (3/7)= -3/7$ or about -0.4286, then $x_2= -0.4181$, etc.

2) x^3 - 2x^2 + x - 3 - cos(2x) = 0
Slope = Rise/Run
Slope: Derivative of x^3 - 2x^2 + x - 3 - cos(2x)
3(1.8)^2 - 4(1.8) + 1 + sin2(1.8)(2)
Rise: Plug in 1.8 to the equation
(1.8)^3-2(1.8)^2 + 1.8 -3 - cos(2(1.8))
Run: 1.8 - x
Final Equation:
3(1.8)^2 - 4(1.8) + 1 + sin2(1.8)(2) = (1.8)^3-2(1.8)^2 + 1.8 -3 - cos(2(1.8)) / 1.8 - x
Simplify: 3.645639188 = -2.846026728 / 1.8 - x
-2.846026728 / 3.645639188 = 1.8 - x
1.8 + 0.780666045 = x
x = 2.680666046
The answer in the book is 2.0923.. So, do you think I should use my "x" to be the second trial and do the whole equation again from the start to get the 2.0923?

I think what you are doing is using the idea behind Newton's method rather than Newton's method itself: Yes, you are trying to solve $f(x)= x^3- 2x^2+ x- 3- cos(2x)= 0$. $f '(x)= 3x^2- 4x+ 1+ 2sin(2x)$. If you choose 1.8 as your x₀, then $x_1= 1.8- ((1.8)^3- 2(1.8)^2+ 1.8- 3- cos(3.6))/(3(1.8)^2- 4(1.8)+ 1+ sin(3.6)= 2.1091$, not the "2.6807" you got. Try it again and then continue until you get two consecutive values that are equal to 4 decimal places.

3)
a = m^2(2m-1)
(0.6543)^2(2(0.6543)-1))
a= 0.13211428
y = a^3
Therefore...
y = 0.13211428^3
y =0.002305946

I'm stuck. Please help?

You weren't asked to find y, you were asked to find the derivative of a: use the chain rule: dy/dm= (dy/da)(da/dm). To find da/dm, I think I would go ahead and multiply: a= m²(2m-1)= m³- m²

4) ((0)($$\sqrt[3]{x+5}$$)) - ((-4)(1/3(x+5)^-2/3)) / ($$\sqrt[3]{x+5}$$)^2)
Simplify: ((4)(1/3(x+5)^-2/3) / ($$\sqrt[3]{x+5}$$)
Then I don't know how to simplify it much longer.. My simplifying skills has gone bad, I guess. Help, please?

I think you will find it much easier to write y= -4(x+ 5)^-1/3 and use the 'power rule' rather than using the quotient rule.

5) Get y-value, plug in the x-value which is 1
1-1 / 1+1 = 0
(1,0)
Do quotient rule on the equation...
(1)(x+1) - (x-1)(1) / (x+1)^2
Simplify...
(x+1) - (x-1) / (x^2 + 2x+2)

No, (x+1)² is NOT "x²+ 2x+ 2! I think you would find it better just to leave the denominator as (x+ 1)².

Plug in the x-value which is 1 to get the slope.
(1+1) - (1-1) / (1^2 +2(1) + 2)
2 / 1+2+2 = 2 / 5
Slope = 2/5

As I said, your denominator is wrong. (1+1)²= 4.

y=mx+b
0 = 2/5(1) + b
0 - 2/5 = b
-2/5 = b
Final equation...
y = 2/5x -2/5

Although it seems like I did everything right, my final equation seems to be incorrect from the answer in the book which is x-2y=1

Yes, see my comments above.

THAT'S ALL FOR NOW. I'll continue more questions later on if I find any that bugs my mind. But for now, I need to go to church so in the meanwhile, PLEASE HELP? =/