1. ## 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

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)

$\displaystyle f'(x) = 3x^2+3*2bx+3c$

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

so

$\displaystyle 3x^2+3*2bx+3c = 0$

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

3. 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) $\displaystyle f(x)=x^3+3bx^2+3cx+d$

$\displaystyle f'(x)=3x^2+6bx+3c$

$\displaystyle x^2+2bx+c=0$

Find the discriminant ...

$\displaystyle (2b)^2-4(1)(c)$

$\displaystyle =4b^2-4c$

$\displaystyle =4(b^2-c)>0$ as we know that $\displaystyle b^2>c$

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