Answer:
Step-by-step explanation
First, let's remember the equation for point-slope form and how to find the 'm' value:
Now plug your m value and (x1, y1) into point-slope form:
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
The percent that is 33 is (33/88)*100 = 37.5%
Answer:
No. It is a constant function.
Step-by-step explanation:
The function f(x) = e^2 is not an exponential functional. Rather, it is a constant function. The reason for this is that in f(x) = e^2, there is no x involved on the right hand side of the equation. The approximate value of e is 2.718281, and the approximate value of 2.718281^2 is 7.389051. This means that f(x) = e^2 = 7.389051. It is important to note that for any value of x, the value of the function remains fixed. This is because the function does not involve the variable x in it. The graph of the function will be a line parallel to the x-axis, and the y-intercept will be 7.389051. For all the lines parallel to x-axis, the value of the function remains the same irrespective of the value of x. Also, the derivative of the function with respect to x is 0, which means that the value of the function is unaffected by the change in the value of x!!!
The answer you are looking for is B. 15
