Find the degree 3 Taylor polynomial T3(x) of the function f(x)=(7x+50)4/3 at a=2
Find the second-degree Taylor polynomial for f(x)=4x2−7x+6 about x=0
thank you! (:
Taylor's series about a (to x² term):
f(x)=f(a) + f'(a)x/1! + f"(a)x²/2!
For
f(x)=(4/3)(7x+50)=(28x/3)+200/3, and a=2
f'(x)=28/3
f"(x)=0
f(x)=f(2) + f'(2)(x-2)/1! + f"(2)(x-2)²/2!
=(4/3)(64) + 28/3(x-2) + 0(x-2)²/2!
=(4/3)(7x-50)
Try the second one along the same lines, and post if you would like to check the answer, which should be the same as the original function.
To find the degree 3 Taylor polynomial, T3(x), of the function f(x) = (7x+50)^(4/3) at a=2, we will need to find the derivatives of the function up to the third degree and evaluate them at a=2.
Step 1: Find f(x) and its derivatives:
f(x) = (7x+50)^(4/3)
Taking the first derivative:
f'(x) = (4/3)(7x + 50)^(1/3) * 7 = 28/3 * (7x + 50)^(1/3)
Taking the second derivative:
f''(x) = (28/3 * (1/3)) * (7x + 50)^(-2/3) * 7 = 28/9 * (7x + 50)^(-2/3)
Taking the third derivative:
f'''(x) = (28/9 * (-2/3)) * (7x + 50)^(-5/3) * 7 = -56/27 * (7x + 50)^(-5/3)
Step 2: Evaluate the derivatives at a=2:
f(2) = (7(2) + 50)^(4/3) = (14 + 50)^(4/3) = 64^(4/3) = 256
f'(2) = 28/3 * (7(2) + 50)^(1/3) = 28/3 * (14 + 50)^(1/3) = 28/3 * 64^(1/3) = 28/3 * 4 = 112/3
f''(2) = 28/9 * (7(2) + 50)^(-2/3) = 28/9 * (14 + 50)^(-2/3) = 28/9 * 64^(-2/3) = 28/9 * 1/4 = 7/9
f'''(2) = -56/27 * (7(2) + 50)^(-5/3) = -56/27 * (14 + 50)^(-5/3) = -56/27 * 64^(-5/3) = -56/27 * 1/32 = -7/27
Step 3: Construct the Taylor polynomial:
The Taylor polynomial, T3(x), is given by:
T3(x) = f(2) + f'(2)(x-2) + (f''(2)/2!)(x-2)^2 + (f'''(2)/3!)(x-2)^3
Substituting the values we calculated:
T3(x) = 256 + (112/3)(x-2) + (7/9)(x-2)^2 + (-7/27)(x-2)^3
Therefore, the degree 3 Taylor polynomial of the function f(x) = (7x+50)^(4/3) at a=2 is:
T3(x) = 256 + (112/3)(x-2) + (7/9)(x-2)^2 + (-7/27)(x-2)^3
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To find the second-degree Taylor polynomial for f(x) = 4x^2 - 7x + 6 about x=0, we will need to find the derivatives of the function up to the second degree and evaluate them at x=0.
Step 1: Find f(x) and its derivatives:
f(x) = 4x^2 - 7x + 6
Taking the first derivative:
f'(x) = 8x - 7
Taking the second derivative:
f''(x) = 8
Step 2: Evaluate the derivatives at x=0:
f(0) = 4(0)^2 - 7(0) + 6 = 6
f'(0) = 8(0) - 7 = -7
f''(0) = 8
Step 3: Construct the Taylor polynomial:
The Taylor polynomial, T2(x), is given by:
T2(x) = f(0) + f'(0)(x-0) + (f''(0)/2!)(x-0)^2
Substituting the values we calculated:
T2(x) = 6 + (-7)(x-0) + (8/2)(x-0)^2
Simplifying:
T2(x) = 6 - 7x + 4x^2
Therefore, the second-degree Taylor polynomial for f(x) = 4x^2 - 7x + 6 about x=0 is:
T2(x) = 6 - 7x + 4x^2
To find the degree 3 Taylor polynomial T3(x) of the function f(x) = (7x + 50)^(4/3) at a = 2, we can use the formula for the Taylor polynomial:
Tn(x) = f(a) + f'(a)(x - a) + (1/2!)f''(a)(x - a)^2 + (1/3!)f'''(a)(x - a)^3 + ... + (1/n!)f^n^(a)(x - a)^n
First, let's find the derivatives of f(x). The first derivative is:
f'(x) = d/dx[(7x + 50)^(4/3)]
= (4/3)(7x + 50)^(1/3) * 7
Evaluating the first derivative at a = 2 gives us:
f'(2) = (4/3)(7(2) + 50)^(1/3) * 7
= 4(57)^(1/3)
Next, we need to find the second derivative:
f''(x) = d/dx[(4/3)(7x + 50)^(1/3)]
= (4/3) * (1/3)(7x + 50)^(-2/3) * 7
Evaluating the second derivative at a = 2:
f''(2) = (4/3) * (1/3)(7(2) + 50)^(-2/3) * 7
= 4(9/157)^(-2/3)
Now, let's find the third derivative:
f'''(x) = d/dx[(4/3)(1/3)(7x + 50)^(-2/3) * 7]
= (4/3) * (1/3) * (-2/3)(7x + 50)^(-5/3) * 7
Evaluating the third derivative at a = 2:
f'''(2) = (4/3) * (1/3) * (-2/3)(7(2) + 50)^(-5/3) * 7
= -8(16/157)^(-5/3)
Now that we have the values for f(2), f'(2), f''(2), and f'''(2), we can substitute them into the formula for the degree 3 Taylor polynomial:
T3(x) = f(2) + f'(2)(x - 2) + (1/2!)f''(2)(x - 2)^2 + (1/3!)f'''(2)(x - 2)^3
Substituting the values and simplifying:
T3(x) = (7(2) + 50)^(4/3) + 4(57)^(1/3)(x - 2) + (1/2!)[4(9/157)^(-2/3)](x - 2)^2 + (1/3!)[-8(16/157)^(-5/3)](x - 2)^3
This gives you the degree 3 Taylor polynomial T3(x) for the function f(x) = (7x + 50)^(4/3) at a = 2.
For the second question:
To find the second-degree Taylor polynomial for f(x) = 4x^2 - 7x + 6 about x = 0, we can use the same formula for the Taylor polynomial:
Tn(x) = f(a) + f'(a)(x - a) + (1/2!)f''(a)(x - a)^2 + ...
First, we need to find the derivatives of f(x). The first derivative is:
f'(x) = d/dx[4x^2 - 7x + 6]
= 8x - 7
The second derivative is:
f''(x) = d/dx[8x - 7]
= 8
Now, let's evaluate each derivative at a = 0:
f(0) = 4(0)^2 - 7(0) + 6
= 6
f'(0) = 8(0) - 7
= -7
f''(0) = 8
Substituting these values into the formula for the second-degree Taylor polynomial:
T2(x) = f(0) + f'(0)(x - 0) + (1/2!)f''(0)(x - 0)^2
Simplifying:
T2(x) = 6 - 7x + (1/2!)(8)(x^2)
This gives you the second-degree Taylor polynomial for f(x) = 4x^2 - 7x + 6 about x = 0.
I hope this helps! Let me know if you have any further questions.