This problem deals with functions defined by f(x) = x^3 - 3bx with b > 0.

[IntegrateMe]

(a) Find the x- and y-coordinates of the relative maximum points of f in terms of b.
(b) Find the x- and y-coordinates of the relative minimum points of f in terms of b.
(c) Show that for all values of b > 0, the relative maximum and minimum points lie on a function of the form y = -ax³ by finding the value of a.

(a)

f(x)=x³-3bx
f'(x)=3x²-3b=0
x²=b
x=+/-sqrt(b)

when x=-sqrt(b),
f(x) = y = -b^3/2 - 3b(-b^1/2)
(x,y) = ( -sqrt(b) , 2b^3/2) f has a maximum

(b)

when x=sqrt(b),
f(x) = y = b^3/2 - 3b(b^1/2)
(x,y)=(sqrt(b),-2b^3/2) f has a minimum

(c)

I'm not sure. Can someone help me with (c)?

Ok, i attempted each part. Did i do anything wrong?

 

[mfb]

(a) and (b) are right. For (c), just use x=sqrt(b) from part (b) and express the y-value in terms of x: y = -2b^3/2 = -2x³. Repeat the same with (a) to cover negative x values, the result will be the same.

 

[nuuskur]

Your terminology is confusing. When we speak of a maximum of $f$, we don't mean a point on the plane, but the value of the function. So, the maximum of $f$ is $f(x_0)$ at the point $x_0$.

For c) assume $b>0$ and repeat what you did before.