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

Question

Charles Arocha · Accepted Answer

Since #f''(x)# 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 Rightarrow 5 ge x#,
so the domain of #f# is #(-infty,5]#. By Product Rule,
#f'(x)=1cdot sqrt{5-x}+xcdot{-1}/{2sqrt{5-x}}={10-3x}/{2sqrt{5-x}}# By Quotient Rule,
#f''(x)={-3cdot2sqrt{5-x}-(10-3x)cdot{-1}/{sqrt{5-x}}}/{4(5-x)}#
#={3x-20}/{4(5-x)^{3/2}}<0# on #(-infty,5)# Hence, #f# is always concave downward.

Emilia Aguilar · Answer

undefined To find the inflection points for the function $ f(x) = x\sqrt{5 - x} $, follow these steps:

1. Find the second derivative of the function $ f(x) $.
2. Set the second derivative equal to zero and solve for $ x $.
3. Determine the corresponding $ y $ values for the $ x $ values found in step 2.

Let's go through these steps:

1. First, find the first derivative of $ f(x) $ using the product rule:

$$ f'(x) = \frac{d}{dx}[x] \cdot \sqrt{5 - x} + x \cdot \frac{d}{dx}[\sqrt{5 - x}] $$

$$ f'(x) = 1 \cdot \sqrt{5 - x} + x \cdot \frac{d}{dx}[\sqrt{5 - x}] $$

$$ f'(x) = \sqrt{5 - x} - \frac{x}{2\sqrt{5 - x}} $$

2. Next, find the second derivative of $ f(x) $:

$$ f''(x) = \frac{d}{dx}\left[\sqrt{5 - x} - \frac{x}{2\sqrt{5 - x}}\right] $$

$$ f''(x) = -\frac{1}{2}\cdot\frac{d}{dx}\left[(5 - x)^{\frac{1}{2}}\right] - \frac{1}{2\sqrt{5 - x}} + \frac{x}{4}\cdot\frac{d}{dx}\left[(5 - x)^{-\frac{1}{2}}\right] $$

$$ f''(x) = -\frac{1}{2}\cdot\left[-\frac{1}{2}(5 - x)^{-\frac{3}{2}}\right] - \frac{1}{2\sqrt{5 - x}} - \frac{x}{4}\cdot\left[-\frac{1}{2}(5 - x)^{-\frac{3}{2}}\right] $$

$$ f''(x) = \frac{1}{4}(5 - x)^{-\frac{3}{2}} - \frac{1}{2\sqrt{5 - x}} + \frac{x}{8}(5 - x)^{-\frac{3}{2}} $$

$$ f''(x) = \frac{1}{4}(5 - x)^{-\frac{3}{2}} + \frac{x}{8}(5 - x)^{-\frac{3}{2}} - \frac{1}{2\sqrt{5 - x}} $$

$$ f''(x) = \frac{1 + 2x - 5x}{8\sqrt{5 - x}(5 - x)} $$

$$ f''(x) = \frac{1 - 3x}{8\sqrt{5 - x}(5 - x)} $$

3. Set $ f''(x) $ equal to zero and solve for $ x $:

$$ \frac{1 - 3x}{8\sqrt{5 - x}(5 - x)} = 0 $$

$$ 1 - 3x = 0 $$

$$ x = \frac{1}{3} $$

4. Determine the corresponding $ y $ value:

$$ f\left(\frac{1}{3}\right) = \frac{1}{3}\sqrt{5 - \frac{1}{3}} = \frac{1}{3}\sqrt{\frac{14}{3}} = \frac{\sqrt{14}}{3} $$

Therefore, the inflection point is $ \left(\frac{1}{3}, \frac{\sqrt{14}}{3}\right) $.