consider the following. (if an answer does not exist, enter dne.)\n\nf(x)=x√(6 - x)\n\n(a) find the…

Explanation:

Step1: Find the domain

For (F(x)=x\sqrt{6 - x}), the expression under the square - root must be non - negative. So (6−x\geq0), which gives (x\leq6). The domain of (F(x)) is ((-\infty,6]).

Step2: Find the first derivative using the product rule

The product rule is ((uv)^\prime = u^\prime v+uv^\prime). Let (u = x) and (v=(6 - x)^{\frac{1}{2}}). Then (u^\prime=1) and (v^\prime=\frac{1}{2}(6 - x)^{-\frac{1}{2}}(-1)=-\frac{1}{2\sqrt{6 - x}}).

(F^\prime(x)=\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 the critical points

Set (F^\prime(x)=0), then (\frac{12 - 3x}{2\sqrt{6 - x}} = 0). Since the denominator (2\sqrt{6 - x}\neq0) for (x<6), we solve (12-3x = 0), which gives (x = 4). The derivative is undefined at (x = 6) (but (x = 6) is an endpoint of the domain).

Step4: Test the intervals for increasing and decreasing

• For the interval ((-\infty,4)), let's choose (x = 0). Then (F^\prime(0)=\frac{12-0}{2\sqrt{6-0}}=\frac{12}{2\sqrt{6}}>0). So (F(x)) is increasing on ((-\infty,4)).
• For the interval ((4,6)), let's choose (x = 5). Then (F^\prime(5)=\frac{12-15}{2\sqrt{6 - 5}}=\frac{-3}{2}<0). So (F(x)) is decreasing on ((4,6)).

Step5: Find local minima and maxima

Since (F(x)) changes from increasing to decreasing at (x = 4), (F(4)=4\sqrt{6 - 4}=4\sqrt{2}) is a local maximum. At (x = 6), (F(6)=6\sqrt{6 - 6}=0). As (F(x)) is decreasing on ((4,6)) and (F(x)) is defined for (x\leq6), (F(6)=0) is a local minimum.

Step6: 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=2\times\frac{1}{2}(6 - x)^{-\frac{1}{2}}(-1)=-(6 - x)^{-\frac{1}{2}})

(F^{\prime\prime}(x)=\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}}})

The second derivative is undefined at (x = 6). Set (F^{\prime\prime}(x)=0), but (3(x - 8)=0) gives (x = 8) which is not in the domain.

Since the second derivative (F^{\prime\prime}(x)<0) for all (x\in(-\infty,6)) (because (x-8<0) and ((6 - x)^{\frac{3}{2}}>0) for (x<6)), the function is concave down on ((-\infty,6)) and there is no inflection point.

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
• Interval where concave up: DNE
• Interval where concave down: ((-\infty,6))