CalculuS derivitives help!!!?

Max & Min points are found by taking first derivative and setting it to zero.

If the second derivative is positive, the point is a minimum...if negative, a maximum...if zero, a point of inflection...as you have noted.

√4-x^2

f ' (x) = 1/2 (4-x^2)^-1/2[-2x]=-x/√4-x^2 = x√4-x^2 / 4-x^2 =0 x=0

x^4+x^3-2x^2-6

f ' (x)=4x^3+3x^2-4x=x(4x^2+3x+4)=0 x=0 x=-3+-√9-64 / 8 ==> √-1 or an imaginary solution

xe^x

f'(x)=e^x+xe^x=e^x(x+1)=0 x=-1

x√16-x^2

f'(x)=√(16-x^2)+1/2(16-x^2)^-1/2 (-2x)=16-x^2[1-2x] / (16-x^2) = 1-2x ==> x=1/2

If you want to find infleciton point you have to take second derivative of the function and then you have to find the roots.In the end if root is change - to + then it is inflection point

you may prepare the chain rule. This function has 5 nested applications: cos, sqrt, sin, tan, and a linear function (?x). undergo in innovations that cos' = -sin, sqrt' = a million / (2*sqrt), sin' = cos, tan' = sec^2 or a million + tan^2, and (ax)' = a y = cos?(sin(tan?x)) y' = -sin?(sin(tan?x)) * (?(sin(tan?x)))' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * (sin(tan?x))' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * cos(tan?x) * (tan?x)' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * cos(tan?x) * (a million + tan^2(?x)) * (?x)' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * cos(tan?x) * (a million + tan^2(?x)) * ? I leave any needed simplication to you, yet I do see a ?(sin(tan?x)) in the two the numerator and denominator.