Find the degree 3 Taylor polynomial T3(x) of the function f(x)=(−3x+33)^4/3 at a=2.
Oh, Taylor polynomials, huh? Well, let me tell you, they're a lot like Taylor Swift songs - they may not be everyone's cup of tea, but they can still get the job done. So let's get our polynomials rolling!
To find the degree 3 Taylor polynomial, T3(x), of the function f(x) = (-3x + 33)^(4/3) at a = 2, we'll need to find the values of the function and its derivatives at that point.
But before we dive in, let me ask you this: why did the polynomial go to the party alone? Because it didn't want any imaginary numbers ruining the fun! Okay, now let's get down to business!
Step 1: Find the first three derivatives of f(x):
The first derivative: f'(x) = 4/3 * (-3x + 33)^(1/3) * (-3)
The second derivative: f''(x) = 4/3 * (-3x + 33)^(-2/3) * (-3)
The third derivative: f'''(x) = (-2/3) * 4/3 * (-3x + 33)^(-5/3) * (-3)
Step 2: Evaluate the function and its derivatives at a = 2:
f(2) = (-3(2) + 33)^(4/3) = 31^(4/3) ≈ 100.932
f'(2) = (4/3 * (-3(2) + 33)^(1/3) * (-3)) = (-3)^(1/3) * (4/3) * (-3) ≈ -7.745
f''(2) = (4/3 * (-3(2) + 33)^(-2/3) * (-3)) = (-3)^(-2/3) * (4/3) * (-3) ≈ 6.117
f'''(2) = (-2/3) * 4/3 * (-3(2) + 33)^(-5/3) * (-3) = (-2/3) * 4/3 * (-3)^(-5/3) * (-3) ≈ 1.632
Step 3: Assemble the Taylor polynomial using the values we've found:
T3(x) = f(2) + f'(2) * (x - 2) + (f''(2) / 2!) * (x - 2)^2 + (f'''(2) / 3!) * (x - 2)^3
Plugging in our values, we get:
T3(x) ≈ 100.932 - 7.745(x - 2) + 6.117/2 * (x - 2)^2 + 1.632/6 * (x - 2)^3
So there you have it, the degree 3 Taylor polynomial T3(x) of the function f(x) = (-3x + 33)^(4/3) at a = 2. Now, go forth and conquer those polynomials like a mathematicial comedian!
To find the degree 3 Taylor polynomial T₃(x) of the function f(x) = (-3x + 33)^(4/3) at a = 2, we will need to calculate the function's derivatives up to the third derivative and evaluate them at x = 2.
Step 1: Find the first derivative of f(x):
f'(x) = d/dx [(-3x + 33)^(4/3)]
Using the chain rule, we get:
f'(x) = (4/3)(-3)(-3x + 33)^(1/3 - 1)
= 4(-3)(-3x + 33)^(-2/3)
Step 2: Find the second derivative of f(x):
f''(x) = d/dx [4(-3)(-3x + 33)^(-2/3)]
We use the chain rule again, giving us:
f''(x) = 4(-3)(-2/3)(-3)(-3x + 33)^(-5/3 - 1)
= 8(-3)(-3)(-2/3)(-3x + 33)^(-8/3)
Step 3: Find the third derivative of f(x):
f'''(x) = d/dx [8(-3)(-3)(-2/3)(-3x + 33)^(-8/3)]
By using the chain rule once more, we obtain:
f'''(x) = 8(-3)(-3)(-2/3)(-8/3)(-3)(-3x + 33)^(-11/3 - 1)
= 2(3^3)(8/3)(8)(-3)(-3x + 33)^(-14/3)
Step 4: Evaluate the derivatives at x = 2:
Evaluate f'(x) at x = 2:
f'(2) = 4(-3)(-3(2) + 33)^(-2/3)
= 4(-3)(27)^(-2/3)
= 4(-3)/(27^(2/3))
= -12/3√27
= -12/9
= -4/3
Evaluate f''(x) at x = 2:
f''(2) = 8(-3)(-3)(-2/3)(-3(2) + 33)^(-8/3)
= 8(-3)(-3)(-2/3)(27)^(-8/3)
= 8(-3)(54)/(27^(8/3))
= 8(-3)(54)/(3^(8/3))
= -1296/3^(3/3)
= -1296/3
= -432
Evaluate f'''(x) at x = 2:
f'''(2) = 2(3^3)(8/3)(8)(-3)(-3(2) + 33)^(-14/3)
= 2(27)(8/3)(8)(-3)(27)^(-14/3)
= 2(27)(8)(-24)/(27^(14/3))
= 2(8)(-24)/(3^(14/3))
= -3072/3^(8/3)
= -3072/27
Step 5: Write the Taylor polynomial T₃(x):
The Taylor polynomial T₃(x) can be written as follows:
T₃(x) = f(a) + f'(a)(x - a) + (f''(a)/2!)(x - a)^2 + (f'''(a)/3!)(x - a)^3
Substituting the values we just calculated, we get:
T₃(x) = f(2) + f'(2)(x - 2) + (f''(2)/2!)(x - 2)^2 + (f'''(2)/3!)(x - 2)^3
= f(2) + (-4/3)(x - 2) + (-432/2)(x - 2)^2 + (-3072/27)(x - 2)^3
Now, we need to find f(2), which is the value of f(x) at x = 2. Substituting x = 2 into the original function f(x), we get:
f(2) = (-3(2) + 33)^(4/3)
= (6 + 33)^(4/3)
= 39^(4/3)
Therefore, the step-by-step construction of the degree 3 Taylor polynomial T₃(x) is:
T₃(x) = 39^(4/3) + (-4/3)(x - 2) + (-432/2)(x - 2)^2 + (-3072/27)(x - 2)^3
To find the degree 3 Taylor polynomial of the function f(x)=(−3x+33)^(4/3) at a=2, we need to first find the coefficients of the polynomial.
The formula for the degree 3 Taylor polynomial of a function f(x) centered at a is given by:
T3(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + (f'''(a)/3!)(x-a)^3
So, let's start by finding the derivatives of f(x):
f(x) = (-3x+33)^(4/3)
f'(x) = (4/3)(-3x+33)^(1/3) * (-3)
= -4(-3x+33)^(1/3)
f''(x) = (4/3) * (-1/3)(-3x+33)^(-2/3) * (-3)
= (4/3)(1/3)(-3x+33)^(-2/3) * (-3)
= 4(-3x+33)^(-2/3)
f'''(x) = (4/3)(-2/3)(-3x+33)^(-5/3) * (-3)
= (4/3)(2/3)(-3x+33)^(-5/3) * 3
= 8(-3x+33)^(-5/3)
Now, we can substitute the values into the formula for the Taylor polynomial:
T3(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + (f'''(a)/3!)(x-a)^3
T3(x) = f(2) + f'(2)(x-2) + (f''(2)/2!)(x-2)^2 + (f'''(2)/3!)(x-2)^3
To calculate the values of f(2), f'(2), f''(2), and f'''(2), substitute x=2 into each of the functions:
f(2) = (-3(2)+33)^(4/3)
= (27)^(4/3)
= 81
f'(2) = 4(-3(2)+33)^(-2/3)
= 4(27)^(-2/3)
= 4/9
f''(2) = 4(-3(2)+33)^(-2/3)
= 4(27)^(-2/3)
= 4/9
f'''(2) = 8(-3(2)+33)^(-5/3)
= 8(27)^(-5/3)
= 8/729
Plugging in these values into the Taylor polynomial formula:
T3(x) = 81 + (4/9)(x-2) + (4/9)/2(x-2)^2 + (8/729)/3!(x-2)^3
Simplifying:
T3(x) = 81 + (4/9)(x-2) + (2/9)(x-2)^2 + (8/2187)(x-2)^3
Therefore, the degree 3 Taylor polynomial, T3(x), for the function f(x)=(−3x+33)^(4/3) at a=2 is:
T3(x) = 81 + (4/9)(x-2) + (2/9)(x-2)^2 + (8/2187)(x-2)^3
f(x) = (-3x+33)^4/3
at x=2, -3x+33 = 27
f(2) = 27^4/3 = 81
f'(x) = -4(-3x+33)^1/3
f'(2) = -4*27^1/3 = -4*3 = -12
f''(x) = 4(-3x+33)^-2/3
f''(2) = 4*27^-2/3 = 4/9
f(3)(x) = 8(-3x+33)^-5/3
f(3)(2) = 8/243
p(x) = f(2) + f'(2)/1! (x-2)^1 + f''(2)/2! (x-2)^2 + f(3)(2)/3! (x-2)^3 + ...
= 81 - 12(x-2) + 2/9 (x-2)^2 + 4/729 (x-2)^3 + ...