Find the Critical Points of a function with square roots and exponents

Question

$$f(x)=\sqrt{x} (x-1)^5$$

I need to find the critical points of this function and I'm really confused. I have found the derivative, but it is very long and I don't know what to do next.

Answer 1

Hint: A critical point is defined to be a point at which the derivative is zero. For example, if $f(x)=x^{2}$, we have $f^{\prime}(x)=2x$. The only critical point of $f$ in this example is $x=0$, since $f^{\prime}(0)=2\cdot0=0$ and for any other $y\neq0$, $f^{\prime}(y)=2y\neq0$.

Answer 2

Here and in the following $x>0$. $$ f'(x) = \frac{(x-1)^5}{2\sqrt{x}} + 5(x-1)^4\sqrt{x} = 0 $$ if and only if (multiplying by $2\sqrt{x}$, which is non-zero) $$ (x-1)^4(x-1 + 10x) = 0 $$ if and only if $x=1$, or $x$ is the root of $11x - 1 = 0$, which I hope you are able to find.

Answer 3

You can get the derivative much faster using logarithmic differentiation $$f(x)=\sqrt{x} (x-1)^5$$ $$\log(f(x))=\frac 12\log(x)+5\log(x-1)$$ $$\frac{f'(x)}{f(x)}=\frac 1{2x}+\frac 5{x-1}$$ $$f'(x)=\Big(\frac 1{2x}+\frac 5{x-1}\Big)f(x)$$ So, the derivative cancels when either $f(x)=0$ or when $\frac 1{2x}+\frac 5{x-1}=0$; after simplification, this leads to a quadratic equation in $x$.