f(x)=x^n , where n is a positive integer greater or equal to 2. The graph of f(x) will have an inflection point when n is
A. Even
B. Odd
C. Divisible by 3
D. For all values
E. For no values
I got B.
It would help you if you say why you chose B.
An inflection point is the point (x,f(x)) such that f"(x) [i.e. second derivative) equals zero.
f(x)=x^n
f'(x)=nx^(n-1)
f"(x)=n(n-1)x^(n-2)
So solve for
f"(x)=0 when n is odd, say n=3, 5, 7...
and solve for
f"(x)=0 when n is even, say n=2,4,6...
and make your conclusion accordingly.
Even? when plug n numbers into formula
To determine when the graph of the function f(x) = x^n will have an inflection point, we need to analyze the behavior of the second derivative of the function.
First, let's find the first and second derivatives of the function f(x):
f'(x) = nx^(n-1)
f''(x) = n(n-1)x^(n-2)
Now, an inflection point occurs when the concavity of the function changes. In other words, the sign of the second derivative changes.
For the second derivative to change sign, we need to consider when f''(x) = n(n-1)x^(n-2) = 0.
Since n is a positive integer greater than or equal to 2, the value of n(n-1) will always be positive. Therefore, f''(x) will never equal zero, and the graph of f(x) = x^n will have no inflection points.
Hence, the correct answer is:
E. For no values