Question

find the intervals on which ( f ) is increasing and the intervals on which it is decreasing.
( f(x)=-x^{2} sqrt{49-x^{2}} ) on ( (-7,7) )
find the intervals of increase of ( f ). select all that apply
a. ( (-7,0) )
b. ( left(\frac{7 sqrt{6}}{3}, 7
ight) )
c. ( left(-\frac{7 sqrt{6}}{7}, \frac{7 sqrt{6}}{7}
ight) )
d. ( left(-\frac{7 sqrt{6}}{3}, 0
ight) )
e. ( left(-7,-\frac{7 sqrt{6}}{3}
ight) )
f. ( left(0, \frac{7 sqrt{6}}{3}
ight) )

Explanation:

Step1: Find the derivative of \(f(x)\)

Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = -x^{2}\), \(u^\prime=-2x\), \(v=\sqrt{49 - x^{2}}=(49 - x^{2})^{\frac{1}{2}}\), and \(v^\prime=\frac{-2x}{2\sqrt{49 - x^{2}}}=\frac{-x}{\sqrt{49 - x^{2}}}\).

\(f^\prime(x)=-2x\sqrt{49 - x^{2}}-x^{2}\cdot\frac{-x}{\sqrt{49 - x^{2}}}=\frac{-2x(49 - x^{2})+x^{3}}{\sqrt{49 - x^{2}}}=\frac{-98x + 2x^{3}+x^{3}}{\sqrt{49 - x^{2}}}=\frac{3x^{3}-98x}{\sqrt{49 - x^{2}}}=\frac{x(3x^{2}-98)}{\sqrt{49 - x^{2}}}\)

Set \(f^\prime(x)=0\), then \(x = 0\) or \(3x^{2}-98=0\), \(x^{2}=\frac{98}{3}\), \(x=\pm\frac{7\sqrt{6}}{3}\) (since \(x\in(-7,7)\))

Step2: Test the intervals

• For the interval \((-7,-\frac{7\sqrt{6}}{3})\), let \(x=- 4\). Then \(f^\prime(-4)=\frac{-4(3\times(-4)^{2}-98)}{\sqrt{49 - (-4)^{2}}}=\frac{-4(48 - 98)}{\sqrt{49 - 16}}=\frac{-4\times(-50)}{\sqrt{33}}>0\)
• For the interval \((-\frac{7\sqrt{6}}{3},0)\), let \(x=-1\). Then \(f^\prime(-1)=\frac{-1(3\times(-1)^{2}-98)}{\sqrt{49 - (-1)^{2}}}=\frac{-1(3 - 98)}{\sqrt{48}}>0\)
• For the interval \((0,\frac{7\sqrt{6}}{3})\), let \(x = 1\). Then \(f^\prime(1)=\frac{1(3\times1^{2}-98)}{\sqrt{49 - 1^{2}}}=\frac{3 - 98}{\sqrt{48}}<0\)

Answer:

C. \((-\frac{7\sqrt{6}}{3},0)\), D. \((-7,-\frac{7\sqrt{6}}{3})\)