Thread: Turning Points

Having real trouble putting these together.

Yor help would be appreciated

Let f(x) = x³ + 3bx² + 3cx + d

a) Show that y =f (x) has two distinct turning points if and only if b² > c


b) If b² > c, show that the vertical distance between the turning points is 4(b² - c) to the power of 3/2

Hi.

Originally Posted by Joel

Having real trouble putting these together.

Yor help would be appreciated

Let f(x) = x³ + 3bx² + 3cx + d

a) Show that y =f (x) has two distinct turning points if and only if b² > c

First you need to find f'(x)



Then you need to find the zeros of f'(x),

so



Now solve for x and you get the turning points.

This is a quadratic equation, you should be able to find them.

Kind regards
Rapha

Originally Posted by Joel

Having real trouble putting these together.

Yor help would be appreciated

Let f(x) = x³ + 3bx² + 3cx + d

a) Show that y =f (x) has two distinct turning points if and only if b² > c


b) If b² > c, show that the vertical distance between the turning points is 4(b² - c) to the power of 3/2

(a)





Find the discriminant ...





as we know that

Hence , y=f(x) has 2 distinct turning points .