Differentiate it
get f ' (x)=3x^2-4
put 3x^2-4=0
x=2/√3,-2/√3
so interval
x∈(-2/√3,2/√3)
Hello :
the normal vector of the plane is : d' = [4,-1,5]...(vector perpendicular to the plane)<span>
d </span><span>⊥ d' because : (4)(2)+(-1)(3)+(5)(1)=0
</span>the line is perpendicular to the plane .
<span>(4x ÷ 2) - 9y
is difference of two terms.</span>
There is nothing there to work with
