Find the critical points of the given function.f(x)=2−x2f(x)=\sqrt{2-x^2}f(x)=2−x2

x = 0 , x = 2 , x = − 2 x=0,x=\sqrt2,x=-\sqrt2 x = 0 , x = 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , x = − 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

Find the critical points of the given function. f ( x ) = 2 − x 2 f(x)=\sqrt{2-x^2} f ( x ) = 2 − x 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z">

In this problem, we're asked to find the critical points of the given function. The function that we're given here is f of x equals the square root of 2 minus x squared. Remember that the critical points are going to be where our derivative, so in this case f prime of x, is either equal to 0 or it does not exist. So let's start here by finding our derivative. Now in finding the derivative, it's going to be easier to think about this as being 2 minus x squared to the power of 1 half so that we can use the power rule here. So finding our derivative f prime of x using the power rule, I'm going to pull that one half to the front. So one half times 2 minus x squared, and then subtracting 1 from that power gives me negative one half and then applying the chain rule, I then need to multiply this by the derivative of what's inside of those parentheses, which is going to be a negative two x. So doing a bit of simplification here, this one half and this 2 are going to cancel and I'm going to go ahead and rewrite that power as a root. So this ends up giving me a negative x on my numerator and then the square root of 2 minus x squared in the denominator. So I wanna find where this derivative is equal to 0 and where it does not exist. Now in setting this equal to 0, remember that since this is a rational function here, I'm effectively just finding where my numerator is equal to 0. So if I set negative x equal to 0, that is really just x equals 0. So that gives me my first critical point there. I also wanna think about where my derivative does not exist. Now because here I have a rational function, my derivative won't exist where this denominator is equal to 0. So if I figure out where this does not exist, by setting that square root of 2 minus x squared equal to 0, that square root will cancel because if I square both sides, I still have 0 on that right side. So that just gives me 2 minus x squared equals 0. Then if I move that x squared to the other side, that gives me 2 equals x squared. Taking the square root of both sides then tells me my final answer that x is equal to plus or minus the square root of 2, having just swapped those sides there. So I have 2 other potential critical points, but remember I need to double check if these values are actually in the original domain of my function. If they're not, then they're not critical points. But if they are, then they are critical points. Now looking at my original function, the square root of 2 minus x squared, if I were to plug in a positive root 2 or negative root 2, I would just end up getting 0 here, and that means that it does exist within the domain of our function. It doesn't give me anything crazy. So these are critical points. And that means that this function has 3 critical points where x is equal to 0, where x is equal to positive root 2, and where x is equal to negative root 2. Now feel free to let us know if you have any questions. Let's keep going.

Find the critical points of the given function. f ( x ) = 2 − x 2 f(x)=\sqrt{2-x^2} f ( x ) = 2 − x 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

x = 0 , x = 2 , x = − 2 x=0,x=\sqrt2,x=-\sqrt2 x = 0 , x = 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , x = − 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

x = 0 , x = 2 x=0,x=\sqrt2 x = 0 , x = 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

6. Graphical Applications of Derivatives

The First Derivative Test

Business Calculus

6. Graphical Applications of Derivatives

The First Derivative Test

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Multiple Choice

Find the critical points of the given function.
$f(x)=\sqrt{2-x^2}$

$x=0$

$x=0,x=\sqrt2,x=-\sqrt2$

$x=0,x=\sqrt2$

No critical points

Verified step by step guidance

Step 1: Recall that critical points occur where the derivative of the function is either zero or undefined. Start by finding the derivative of the given function f(x) = √(2 - x²).

Step 2: Use the chain rule to differentiate f(x). The derivative of √(2 - x²) is (1/2)(2 - x²)^(-1/2) multiplied by the derivative of the inner function (2 - x²), which is -2x.

Step 3: Simplify the derivative. The result is f'(x) = -x / √(2 - x²).

Step 4: Set the derivative equal to zero to find where the slope of the tangent line is zero. Solve -x / √(2 - x²) = 0. This occurs when x = 0.

Step 5: Determine where the derivative is undefined. The derivative is undefined when the denominator √(2 - x²) = 0, which happens when 2 - x² = 0. Solve for x to find x = ±√2. Therefore, the critical points are x = 0, x = √2, and x = -√2.

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