# Turning Points

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• June 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
• June 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)

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

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

so

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

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
• June 14th 2009, 11:54 PM
mathaddict
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) $f(x)=x^3+3bx^2+3cx+d$

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

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

Find the discriminant ...

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

$=4b^2-4c$

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

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