Determine all critical points for the function.

f(x) = (x - 2)7

Question

Determine all critical points for the function.

f(x) = (x - 2)7

BlackTom AI · Accepted Answer

We start by identifying all critical points of f(x) = (x - 2)^7.
Option 1: x = 0 and x = 2. To evaluate this, we should consider where the derivative is zero or undefined. The derivative f'(x) = 7(x - 2)^6, which is zero only when x = 2. The value x = 0 does not make f'(x) zero, so this option is incorrect.
Option 2: x = 2. This matches the point where the derivative 7(x - 2)^6 equals zero, since any even power of (x - 2) vanishes at x = 2. There are no other x-values giving f'(x) = 0, so this is a candidate for a critical point.
Option 3: x = 2 and x = 7. As established, the derivative vanishes only at x = 2; x = 7 does not satisfy f'(x) = 0, so this option incorrectly adds an extraneous point.
Option 4: x = 0, x = 2, and x = 7. This option lists three points, but only x = 2 makes f'(x) zero, while x = 0 and x = 7 do not. Hence this option is not correct.
Additionally, recall that critical points occur where the derivative is zero or undefined. Since f is a polynomial, f' is defined for all real x, so there are no endpoints or undefined points to consider here. The only potential critical points come from solving f'(x) = 0, which yields x = 2.
Putting these observations together, only the candidate x = 2 satisfies the critical-point condition via f'(x) = 0, and the other options fail because they either include extra points or omit the true one.

Determine all critical points for the function. f(x) = (x - 2)7 单项选择题

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