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

Just need help with part B

SpaceTsunaml Mar 12, 2021

#1**+2 **

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

ElectricPavlov Mar 12, 2021

#2**0 **

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