How do you find the intervals of increasing and decreasing using the first derivative given #y=2xsqrt(9-x^2)#?

Answer 1

Increasing: #x: (-(3sqrt2)/2, +(3sqrt2)/2)#
Decreasing: #x: [-3, -(3sqrt2)/2) uu (+(3sqrt2)/2, +3]#

#y= 2xsqrt(9-x^2)#
First let's find the domain of #y#
#y in RR# where #(9-x^2)>= 0 -> absx<= 3#
Hence the domain of #y# is: #-3<=x<=+3#
Next we will find the turning points of #y# using #y'=0#
#y = 2x(9-x^2)^(1/2)#

Applying the product rule and chain rule

#y'# = #2x*1/2*(9-x^2)^(-1/2) * (-2x) + 2*(9-x^2)^(1/2)#
#= 2(9-x^2)^(1/2) - (2x^2)/((9-x^2)^(1/2))#
#= (2(9-x^2)-2x^2)/(sqrt(9-x^2))#
#= (18-4x^2)/sqrt(9-x^2)#
For critical #y#:
# (18-4x^2)/sqrt(9-x^2)=0 -> 18-4x^2=0#
#2x^2 =9 -> x= +- sqrt(9/2)#
#x= +-3/sqrt2 = +-(3sqrt2)/2#
This question asks for the intervals of #x# of increasing and decreasing #y# using the #y'#. To avoid using the second derivative, it is now helpful to observe the graph of #y# below:

graph{2xsqrt(9-x^2) [-20.28, 20.26, -10.13, 10.15]}

Since we know the turning points are where #x= +-(3sqrt2)/2# and that the domain of #y# is #x: [-3, +3]#
We can see that #y# is increasing for #x: (-(3sqrt2)/2, +(3sqrt2)/2)#
and #y# is decresing for #x: [-3, -(3sqrt2)/2) uu (+(3sqrt2)/2, +3]#
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Answer 2

To find the intervals of increasing and decreasing using the first derivative:

  1. Find the first derivative of the function y=2x√(9-x^2).
  2. Set the first derivative equal to zero and solve for x to find critical points.
  3. Determine the sign of the first derivative in each interval created by the critical points.
  4. If the first derivative is positive in an interval, the function is increasing in that interval. If it is negative, the function is decreasing.

Let's proceed with these steps:

  1. The given function is (y=2x\sqrt{9-x^2}). First, find the first derivative, (y'): [y' = \frac{d}{dx} \left(2x\sqrt{9-x^2}\right)]

  2. Next, simplify and differentiate using the product rule and chain rule: [y' = 2\sqrt{9-x^2} + 2x \left(\frac{1}{2\sqrt{9-x^2}}\right) (-2x)] [y' = 2\sqrt{9-x^2} - \frac{2x^2}{\sqrt{9-x^2}}]

  3. Set (y') equal to zero and solve for (x) to find critical points: [2\sqrt{9-x^2} - \frac{2x^2}{\sqrt{9-x^2}} = 0]

  4. Solve for (x) to find critical points.

    [2\sqrt{9-x^2} = \frac{2x^2}{\sqrt{9-x^2}}] [2(9-x^2) = x^2] [18 - 2x^2 = x^2] [3x^2 = 18] [x^2 = 6] [x = \pm \sqrt{6}]

    The critical points are (x = -\sqrt{6}) and (x = \sqrt{6}).

  5. Determine the sign of (y') in each interval:

    Test a value less than (-\sqrt{6}) (e.g., (x = -\sqrt{7})): [2\sqrt{9-(-\sqrt{7})^2} - \frac{2(-\sqrt{7})^2}{\sqrt{9-(-\sqrt{7})^2}} = 2\sqrt{9-7} - \frac{2(7)}{\sqrt{9-7}}] Since (\sqrt{9-7}) is positive, the first term is positive. Since (\sqrt{9-7}) is also positive, the second term is negative. Thus, (y') is negative for (x < -\sqrt{6}).

    Test a value between (-\sqrt{6}) and (\sqrt{6}) (e.g., (x = 0)): [2\sqrt{9-0^2} - \frac{2(0)^2}{\sqrt{9-0^2}} = 2\sqrt{9} - \frac{0}{\sqrt{9}}] Both terms are positive. Thus, (y') is positive for (-\sqrt{6} < x < \sqrt{6}).

    Test a value greater than (\sqrt{6}) (e.g., (x = \sqrt{7})): [2\sqrt{9-\sqrt{7}^2} - \frac{2(\sqrt{7})^2}{\sqrt{9-\sqrt{7}^2}} = 2\sqrt{9-7} - \frac{2(7)}{\sqrt{9-7}}] Since (\sqrt{9-7}) is positive, the first term is positive. Since (\sqrt{9-7}) is also positive, the second term is negative. Thus, (y') is negative for (x > \sqrt{6}).

  6. Summarizing the intervals:

    • (y') is negative for (x < -\sqrt{6}), so the function (y) is decreasing on ((-\infty, -\sqrt{6})).
    • (y') is positive for (-\sqrt{6} < x < \sqrt{6}), so the function (y) is increasing on ((- \sqrt{6}, \sqrt{6})).
    • (y') is negative for (x > \sqrt{6}), so the function (y) is decreasing on ((\sqrt{6}, \infty)).

Therefore, the intervals of increasing and decreasing are ((- \sqrt{6}, \sqrt{6})) and ((-\infty, -\sqrt{6}) \cup (\sqrt{6}, \infty)), respectively.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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