consider the following. (if an answer does not exist, enter dne.)\n\n$f(x)=x\\sqrt{6 - x}$\n\n(a) find the…

Explanation:

Step1: Find the domain

For (y = x\sqrt{6 - x}), the domain is (x\leqslant6) since the expression under the square - root must be non - negative ((6 - x\geqslant0)).

Step2: Find the first derivative using the product rule

The product rule is ((uv)^\prime=u^\prime v + uv^\prime), where (u = x) and (v=(6 - x)^{\frac{1}{2}}). (u^\prime = 1) and (v^\prime=\frac{1}{2}(6 - x)^{-\frac{1}{2}}(- 1)=-\frac{1}{2\sqrt{6 - x}}) (y^\prime=\sqrt{6 - x}+x\left(-\frac{1}{2\sqrt{6 - x}}\right)=\frac{2(6 - x)-x}{2\sqrt{6 - x}}=\frac{12-2x - x}{2\sqrt{6 - x}}=\frac{12 - 3x}{2\sqrt{6 - x}})

Step3: Find critical points

Set (y^\prime = 0), then (12-3x = 0), so (x = 4). The derivative is undefined at (x = 6) (but (x = 6) is the endpoint of the domain).

• Interval of increase: Test the interval ((-\infty,4)). Let (x = 0), then (y^\prime=\frac{12-0}{2\sqrt{6-0}}=\frac{12}{2\sqrt{6}}>0). So the function is increasing on ((-\infty,4))
• Interval of decrease: Test the interval ((4,6)). Let (x = 5), then (y^\prime=\frac{12-15}{2\sqrt{6 - 5}}=\frac{-3}{2}<0). So the function is decreasing on ((4,6))

Step4: Find local extrema

Since the function changes from increasing to decreasing at (x = 4), the local maximum value is (y(4)=4\sqrt{6 - 4}=4\sqrt{2}). At (x = 6), (y(6)=6\sqrt{6 - 6}=0). Since the function is decreasing on ((4,6)) and the domain is (x\leqslant6), the local minimum value is (0)

Step5: Find the second derivative

Using the quotient rule (\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = 12-3x) and (v = 2(6 - x)^{\frac{1}{2}}) (u^\prime=-3) and (v^\prime=(6 - x)^{-\frac{1}{2}}(-1)) (y^{\prime\prime}=\frac{-3\times2\sqrt{6 - x}-(12 - 3x)\left(-\frac{1}{\sqrt{6 - x}}\right)}{4(6 - x)}=\frac{-6(6 - x)+(12 - 3x)}{4(6 - x)^{\frac{3}{2}}}=\frac{-36 + 6x+12 - 3x}{4(6 - x)^{\frac{3}{2}}}=\frac{3x - 24}{4(6 - x)^{\frac{3}{2}}}=\frac{3(x - 8)}{4(6 - x)^{\frac{3}{2}}})

Step6: Find inflection points and concavity

Set (y^{\prime\prime}=0), the numerator (3(x - 8)=0) gives (x = 8) (but (x = 8) is not in the domain). The second derivative is undefined at (x = 6). Test the interval ((-\infty,6)). Let (x = 0), (y^{\prime\prime}=\frac{3(0 - 8)}{4(6 - 0)^{\frac{3}{2}}}=\frac{-24}{4\times6\sqrt{6}}<0). So the function is concave down on ((-\infty,6)) and there is no inflection point (since there is no change in concavity within the domain)

Answer:

(a) Interval of increase: ((-\infty,4)); Interval of decrease: ((4,6)) (b) Local minimum value: (0); Local maximum value: (4\sqrt{2}) (c) Inflection point: DNE; Concave up: DNE; Concave down: ((-\infty,6))