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I got part A, as \(x = {}\frac{1-2x}{\sqrt{1-x^2}}\)

Just need help with part B

 Mar 12, 2021
edited by SpaceTsunaml  Mar 12, 2021
 #1
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So your question is looking for where the tangent line has zero slope in the interval between  -1 and 1  inclusive

 

  If you properly derived the derivative,  set it equal  to 0   and find where in the inteval requested, it = zero

     The derivative of a funtion IS the slope of the function     ....... given f(x)     f ' (x) is the slope.....(  gradient is the same thing as slope.)

 

 

1-2x / (sqrt1-x^2)    = 0

1-2x = 0

-2x = -1

x = 1/2

 Mar 12, 2021
edited by ElectricPavlov  Mar 12, 2021
 #2
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See I thought so, but my brain was thinking the gradient OF the derivative, but I felt that was wrong, but anyways yeah thats what I thought haha, it just seemed too simple! Lol my brain overthinking everything. Thank you :)

SpaceTsunaml  Mar 12, 2021

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