2Qs Local min and area

[davet]

Hi everyone, I am having trouble with two calculus problems. First

1) The function F(x) = arcsinex - square root of 2 multiplied by x :domain is [-1,1] has a local min at x =?

Note: the X is not under the square root. Derivative of arcsine is 1/square root of 1-x^2 but the square root 2 multiplied by x is the trouble I'm having. Finding the critical points is kinda difficult for me in this type of problem. I tried using the product rule for the square root 2 times x but have gotten something that looks extremely diffcult to handle.


2) The area of the region bounded by the curves y=x^2, y = 3-2x and y = 0.

At first I thought area between curves was an easy problem but this one is giving me trouble. I bascially found the boundaries of -3 and 1 by setting the 2 equations equal to each other. Then I did the top curve - the bottom curve and took the antiderivative of it obtaining 3x -x^2 -x^3/3 with boundaries -3 and 1. The answer is 7/12, but I cannot obtain a denominator other than 3. This is confusing. Hope someone can help, Thanks in advance.

 

[galactus]

Hi everyone, I am having trouble with two calculus problems. First

1) The function F(x) = arcsinex - square root of 2 multiplied by x :domain is [-1,1] has a local min at x =?

Note: the X is not under the square root. Derivative of arcsine is 1/square root of 1-x^2 but the square root 2 multiplied by x is the trouble I'm having. Finding the critical points is kinda difficult for me in this type of problem. I tried using the product rule for the square root 2 times x but have gotten something that looks extremely diffcult to handle.

\(\displaystyle \L\\\frac{d}{dx}[sin^{-1}(x)-x\sqrt{2}]=\frac{1}{\sqrt{1-x^{2}}}-\sqrt{2}\)

\(\displaystyle \L\\\frac{1}{\sqrt{1-x^{2}}}-\sqrt{2}=0\)

\(\displaystyle \L\\x=\pm\frac{1}{\sqrt{2}}\)

\(\displaystyle f''(x)=\frac{-x}{(x^{2}-1)\sqrt{1-x^{2}}}\)

Subbing \(\displaystyle \frac{1}{\sqrt{2}}\) into f''(x) we get 2. Which means it's a relative minimum.

2) The area of the region bounded by the curves y=x^2, y = 3-2x and y = 0.

At first I thought area between curves was an easy problem but this one is giving me trouble. I bascially found the boundaries of -3 and 1 by setting the 2 equations equal to each other. Then I did the top curve - the bottom curve and took the antiderivative of it obtaining 3x -x^2 -x^3/3 with boundaries -3 and 1. The answer is 7/12, but I cannot obtain a denominator other than 3. This is confusing. Hope someone can help, Thanks in advance.

The answer to this is not 7/12. It's 32/3.

\(\displaystyle \L\\\int_{-3}^{1}(3-2x-x^{2})dx=\frac{32}{3}\)

 

[davet]

You are awesome, thank you. I knew the second problem was somehow wrong. I'll have to email my prof about it. Thanks again.