# Turning Points

• Jun 14th 2009, 11:21 PM
Joel
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
• Jun 14th 2009, 11:50 PM
Rapha
Hi.

Quote:

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
• Jun 14th 2009, 11:54 PM
Quote:

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 .