# Learning Calculus - Finding turning points.

• December 5th 2011, 04:52 AM
Rastanking
Learning Calculus - Finding turning points.
Hi everyone,

I'm trying to teach myself calculus and have come across this problem.

I have a function of x

f(x) = x^3 + 5x^2 - 8x - 12

so I set this to zero by

f'(x) = 3x^2 + 10x - 8 = 0

then the book I'm using says that I can find the local values of x for min and max and these should be x1 = 0.667 and x2 = -4.000 but it doesn't say how these figures are reached.

Could someone guide me in the right direction please.

Andy
• December 5th 2011, 05:00 AM
e^(i*pi)
Re: Learning Calculus
Quote:

Originally Posted by Rastanking
Hi everyone,

I'm trying to teach myself calculus and have come across this problem.

I have a function of x

f(x) = x^3 + 5x^2 - 8x - 12

so I set this to zero by

f'(x) = 3x^2 + 10x - 8 = 0

then the book I'm using says that I can find the local values of x for min and max and these should be x1 = 0.667 and x2 = -4.000 but it doesn't say how these figures are reached.

Could someone guide me in the right direction please.

Andy

By solving $f'(x) = 0$ - in this case using your favourite method for solving quadratic equations
• December 5th 2011, 05:07 AM
Re: Learning Calculus
Quote:

Originally Posted by Rastanking
Hi everyone,

I'm trying to teach myself calculus and have come across this problem.

I have a function of x

f(x) = x^3 + 5x^2 - 8x - 12

so I set this to zero by

f'(x) = 3x^2 + 10x - 8 = 0

then the book I'm using says that I can find the local values of x for min and max and these should be x1 = 0.667 and x2 = -4.000 but it doesn't say how these figures are reached.

Could someone guide me in the right direction please.

Andy

When you have a quadratic equation of the form

$ax^2+bx+c=0$

The "x" solutions are simple to see if you can write your sum of 3 terms as a product by factoring to

$(3x+?)(x+?)=0$

We take the factors of -8 that sum to 10 when multiplying out the brackets.
This is (3)4+1(-2) and so

$3x^2+10x-8=0\Rightarrow\ (3x-2)(x+4)=0$

Since 0(anything)=0

then those factors are giving you the 2 solutions

$3x-2=0$

$x+4=0$

Alternatively,

for $ax^2+bx+c=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
• December 5th 2011, 05:30 AM
Rastanking
Re: Learning Calculus
Thanks for the quick reply, it has been a great help. However the answers in the book are 0.667 and -4.000 but I end up with -1.33 and -2.

I'm either losing the plot entirely or having finger trouble, could someone possible check my maths please.
• December 5th 2011, 05:51 AM
Re: Learning Calculus
You should show how you are getting those answers.
Are you factoring or using the "quadratic formula" ?
• December 5th 2011, 06:01 AM
Rastanking
Re: Learning Calculus
I'm using the quadratic formula and ending up with -1.33 and -2
• December 5th 2011, 06:30 AM
Re: Learning Calculus
$f'(x)=0\Rightarrow\ 3x^2+10x-8=0$

$\Rightarrow\ a=3,\;\;b=10,\;\;c=-8$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-10\pm\sqrt{100-4(3)(-8)}}{6}$

$x=\frac{-10\pm\sqrt{196}}{6}$

$x_1=\frac{\sqrt{196}-10}{6}$

$x_2=\frac{-10-\sqrt{196}}{6}$

One of these is positive and the other is negative.

Your error was to take c=8 instead of minus 8.

$ax^2+bx+c=(3)x^2+(10)x+(-8)$
• December 5th 2011, 10:35 AM
Rastanking
Re: Learning Calculus
Many thanks again for your help, very much appreciated.