curve sketching question: y = x * sqrt[9 - x^2]: find intervals of increase/decrease, max/mins, etc

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coooool222

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Jun 1, 2020
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I am super confused with this question, I'm not able to find the right critical points. My work isn't correlating with the graph.
1680023466697.png

Here is my work trying to find the critical points
[math]f(x) = x{\sqrt {9-x^2}}[/math][math]f'(x) = \frac{-2x^2-x+18}{2{\sqrt{9-x^2}}}[/math][math]-2x^2-x+18=0[/math][math] x = 2.76???? [/math]
I am getting x = 2.76 for the critical points but as you can clearly see on the graph it's +- 2.112?????????
1680037947181.png
I really need help with this thanks
 
Your derivative is wrong. Please try again. After you find the derivative using the product rule, then I'll show you a way without using the product rule.

Also, I am sure that those numbers you put were all approximate. You need exact answers.
 
coooool222 said:
I am super confused with this question, I'm not able to find the right critical points. My work isn't correlating with the graph.
View attachment 35371

Here is my work trying to find the critical points
[math]f(x) = x{\sqrt {9-x^2}}[/math][math]f'(x) = \frac{-2x^2-x+18}{2{\sqrt{9-x^2}}}[/math][math]-2x^2-x+18=0[/math][math] x = 2.76???? [/math]
I am getting x = 2.76 for the critical points but as you can clearly see on the graph it's +- 2.112?????????
View attachment 35374
I am getting x = 2.76 for the critical points but as you can clearly see on the graph it's +- 2.112?????????
Click to expand...
How can you easily tell that the x value is 2.112. Your eye sight is not that good-sorry.
 
Critical point are not just values that make a derivative = 0 but also when .......
 
Last edited:
Steven G said:
Your derivative is wrong. Please try again. After you find the derivative using the product rule, then I'll show you a way without using the product rule.

Also, I am sure that those numbers you put were all approximate. You need exact answers.
Click to expand...
Oh man i forgot the chain rule thanks ???
 
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