Find critical numbers of f(x) = x^5 - 5x

Differentiate: $f'(x) = 5x^4 - 5 = 5(x^4 - 1)$. This is a polynomial, so it is defined everywhere, meaning the only critical numbers come from setting $f'(x) = 0$.

Solve $5(x^4-1) = 0 \Rightarrow x^4 = 1 \Rightarrow x = \pm 1$ (the equation $x^2 = -1$ gives no real solutions). So the critical numbers are $x = -1$ and $x = 1$.

The critical numbers $x=-1$ and $x=1$ split the real line into three intervals: $(-\infty,-1)$, $(-1,1)$, and $(1,\infty)$. Test a sample point in each: at $x=-2$, $f'(-2)=5(16-1)=75>0$; at $x=0$, $f'(0)=5(0-1)=-5<0$; at $x=2$, $f'(2)=5(16-1)=75>0$.

So $f'(x)>0$ on $(-\infty,-1)$ and $(1,\infty)$, meaning $f$ is increasing there, and $f'(x)<0$ on $(-1,1)$, meaning $f$ is decreasing there.

At $x=-1$, $f'(x)$ changes from positive (on $(-\infty,-1)$) to negative (on $(-1,1)$). By the First Derivative Test, this sign change from $+$ to $-$ indicates a local maximum at $x=-1$, with value $f(-1) = -1+5 = 4$.

At $x=1$, $f'(x)$ changes from negative (on $(-1,1)$) to positive (on $(1,\infty)$). This sign change from $-$ to $+$ indicates a local minimum at $x=1$, with value $f(1) = 1-5 = -4$.

The function is decreasing between the two critical numbers, negative one and one. This matches the earlier sign analysis, where a test point like zero gave a negative derivative value. Since f prime is negative on that interval, f is decreasing there, consistent with the local maximum at negative one and local minimum at one.