How do you find the intervals of increasing and decreasing using the first derivative given #y=xsqrt(16-x^2)#?

The function is #y=xsqrt(16-x^2)#

We need the domain of #y#

#16-x^2>=0#, #=>#, #x^2>=16#

Therefore the domain is #x in (-4,4)#

#(uv)'=u'v+uv'#

#u(x)=x#, #=>#, #u'(x)=1#

#v(x)=sqrt(16-x^2)#, #=>#, #v'(x)=(-2x)/(2sqrt(16-x^2))=-x/(sqrt(16-x^2))#

#dy/dx=sqrt(16-x^2)-x^2/(sqrt(16-x^2))=(16-x^2-x^2)/(sqrt(16-x^2))#

#=(16-2x^2)/(sqrt(16-x^2))#

The critical points are when #dy/dx=0#

#(16-2x^2)/(sqrt(16-x^2))=0#

#16-x^2=0#, #=>#, #x^2=8#, #=>#, #x=+-sqrt8#

#color(white)(aaaa)##Interval##color(white)(aaaa)##(-4,-sqrt8)##color(white)(aaaa)##(-sqrt8, sqrt8)##color(white)(aaaa)##(sqrt8,4)#

#color(white)(aaaa)##dy/dx##color(white)(aaaaaaaaaaaaa)##-##color(white)(aaaaaaaaaaaa)##+##color(white)(aaaaaaaaa)##-#

#color(white)(aaaa)##y##color(white)(aaaaaaaaaaaaaaa)##↘##color(white)(aaaaaaaaaaaa)##↗##color(white)(aaaaaaaaa)##↘#

User is studying mathematics and is interested in learning about calculus.To find the intervals of increasing and decreasing for the function \( y = x \sqrt{16 - x^2} \), we need to use the first derivative test. 1. Find the first derivative \( y' \) of the function. 2. Set \( y' = 0 \) and solve for \( x \) to find the critical points. 3. Determine the sign of \( y' \) in each interval created by the critical points. - If \( y' > 0 \), the function is increasing. - If \( y' < 0 \), the function is decreasing. 4. Create an interval table to summarize the information. Let's start by finding the first derivative: \[ y = x \sqrt{16 - x^2} \] \[ y' = \frac{d}{dx} \left( x \sqrt{16 - x^2} \right) \] \[ y' = \sqrt{16 - x^2} + \frac{x(-2x)}{2\sqrt{16 - x^2}} \] \[ y' = \sqrt{16 - x^2} - \frac{x^2}{\sqrt{16 - x^2}} \] Setting \( y' \) to zero and solving for \( x \) gives us the critical points. However, it's worth noting that the function is only defined for \( -4 \leq x \leq 4 \) due to the square root term. \[ \sqrt{16 - x^2} - \frac{x^2}{\sqrt{16 - x^2}} = 0 \] \[ \sqrt{16 - x^2} = \frac{x^2}{\sqrt{16 - x^2}} \] \[ 16 - x^2 = x^2 \] \[ 2x^2 = 16 \] \[ x^2 = 8 \] \[ x = \pm 2\sqrt{2} \] The critical points are \( x = -2\sqrt{2} \) and \( x = 2\sqrt{2} \). Since the function is not defined outside the interval \([-4, 4]\), we can ignore \( x = 2\sqrt{2} \) as a critical point. Now, we test the intervals \( x < -2\sqrt{2} \), \(-2\sqrt{2} < x < 2\sqrt{2}\), and \( x > 2\sqrt{2} \) by plugging in test points into \( y' \) to determine the sign: For \( x < -2\sqrt{2} \), choose \( x = -3 \): \[ y' = \sqrt{16 - (-3)^2} - \frac{(-3)^2}{\sqrt{16 - (-3)^2}} \] \[ y' = \sqrt{7} - \frac{9}{\sqrt{7}} \] \[ y' < 0 \] For \(-2\sqrt{2} < x < 2\sqrt{2}\), choose \( x = 0 \): \[ y' = \sqrt{16 - 0^2} - \frac{0^2}{\sqrt{16 - 0^2}} \] \[ y' = 4 - 0 \] \[ y' > 0 \] For \( x > 2\sqrt{2} \), choose \( x = 3 \): \[ y' = \sqrt{16 - 3^2} - \frac{3^2}{\sqrt{16 - 3^2}} \] \[ y' = \sqrt{7} - \frac{9}{\sqrt{7}} \] \[ y' < 0 \] Therefore, the function is increasing on the interval \( -2\sqrt{2} < x < 2\sqrt{2} \) and decreasing on the intervals \( x < -2\sqrt{2} \) and \( x > 2\sqrt{2} \).