# How do you find the inflection points for the function f(x)=xsqrt(5-x)?

Sep 12, 2014

Since $f ' ' \left(x\right)$ is always negative in the domain of $f$, the graph of $f$ is always concave downward; therefore, there is no inflection point.

Let us first find the domain of $f$.
Since the expression inside the square-root cannot be negative,
$5 - x \ge 0 R i g h t a r r o w 5 \ge x$,
so the domain of $f$ is $\left(- \infty , 5\right]$.

By Product Rule,
$f ' \left(x\right) = 1 \cdot \sqrt{5 - x} + x \cdot \frac{- 1}{2 \sqrt{5 - x}} = \frac{10 - 3 x}{2 \sqrt{5 - x}}$

By Quotient Rule,
$f ' ' \left(x\right) = \frac{- 3 \cdot 2 \sqrt{5 - x} - \left(10 - 3 x\right) \cdot \frac{- 1}{\sqrt{5 - x}}}{4 \left(5 - x\right)}$
$= \frac{3 x - 20}{4 {\left(5 - x\right)}^{\frac{3}{2}}} < 0$ on $\left(- \infty , 5\right)$

Hence, $f$ is always concave downward.