find the critical points of f. assume a is a constant.\n\nf(x)=\\frac{1}{15}x^{15}-a^{14}x\n\nselect the…

find the critical points of f. assume a is a constant.\n\nf(x)=\\frac{1}{15}x^{15}-a^{14}x\n\nselect the correct choice below and fill in any answer boxes within your choice.\n\na. x=\\square\n(use a comma to separate answers as needed.)\n\nb. f has no critical points.

find the critical points of f. assume a is a constant.\n\nf(x)=\\frac{1}{15}x^{15}-a^{14}x\n\nselect the correct choice below and fill in any answer boxes within your choice.\n\na. x=\\square\n(use a comma to separate answers as needed.)\n\nb. f has no critical points.

Answer

Explanation:

Step1: Differentiate the function

According to the power rule ((x^n)^\prime=nx^{n - 1}), if (y = f(x)=\frac{1}{15}x^{15}-a^{14}x), then (f^\prime(x)=\frac{1}{15}\times15x^{14}-a^{14}). Simplify the derivative: (f^\prime(x)=x^{14}-a^{14}).

Step2: Set the derivative equal to zero

Set (f^\prime(x) = 0), so (x^{14}-a^{14}=0). Using the difference - of - powers formula (A^n - B^n=(A - B)(A^{n - 1}+A^{n - 2}B+\cdots+AB^{n - 2}+B^{n - 1})) (here (n = 14), (A=x), (B = a)), we have ((x - a)(x^{13}+x^{12}a+\cdots+xa^{12}+a^{13})=0). Since (x^{14}-a^{14}=(x^7 - a^7)(x^7 + a^7)) and (x^7 - a^7=(x - a)(x^6+ax^5+a^2x^4+a^3x^3+a^4x^2+a^5x + a^6)), (x^7 + a^7=(x + a)(x^6-ax^5+a^2x^4-a^3x^3+a^4x^2-a^5x + a^6)) for (n = 7). The solutions of (x^{14}-a^{14}=0) are (x=a) and (x=-a).

Answer:

A. (x = a,-a)