Answer:
Step-by-step explanation:
correct equation is n - 31 = 52
Answer:
Σ(-1)^kx^k for k = 0 to n
Step-by-step explanation:
The nth Maclaurin polynomials for f to be
Pn(x) = f(0) + f'(0)x + f''(0)x²/2! + f"'(0)x³/3! +. ......
The given function is.
f(x) = 1/(1+x)
Differentiate four times with respect to x
f(x) = 1/(1+x)
f'(x) = -1/(1+x)²
f''(x) = 2/(1+x)³
f'''(x) = -6/(1+x)⁴
f''''(x) = 24/(1+x)^5
To calculate with a coefficient of 1
f(0) = 1
f'(0) = -1
f''(0) = 2
f'''(0) = -6
f''''(0) = 24
Findinf Pn(x) for n = 0 to 4.
Po(x) = 1
P1(x) = 1 - x
P2(x) = 1 - x + x²
P3(x) = 1 - x+ x² - x³
P4(x) = 1 - x+ x² - x³+ x⁴
Hence, the nth Maclaurin polynomials is
1 - x+ x² - x³+ x⁴ +.......+(-1)^nx^n
= Σ(-1)^kx^k for k = 0 to n
Answer:
Step-by-step explanation:
It would take Johny 28 mins to type a 2,100 word essay.
For this parabola we have:
f ( 0 ) = 8
and : f ( 1 ) = 24
In the first equation ( A) :
f ( 0 ) = - 16 * ( 0 - 1 )² + 24 = - 16 * 1 + 24 = 8 ( correct )
f ( 1 ) = - 16 * ( 1 - 1 )² + 24 = 24 ( correct )
For B:
f ( 0 ) = - 16 * ( 0 + 1 )² + 24 = - 16 + 24 = 8 ( correct )
f ( 1 ) = - 16 * ( 1 + 1 )² + 24 = - 16 * 4 + 24 = - 64 + 24 = 40 ( false )
For C:
f ( 0 ) = - 16 * ( 0 - 1 )² - 24 = - 16 - 24 = - 40 ( false )
f ( 1 ) = - 16 * ( 1 - 1 )² - 24 = - 24 ( false )
For D:
f ( 0 ) = - 16 * ( 0 + 1 )² - 24 = - 16 - 24 = - 40 ( false )
f ( 1 ) = - 16 * ( 1 - 1 )² - 24 = - 24 ( false )
Answer:
A ) f ( t ) = - 16 * ( t - 1 )² + 24
