How to find Maximum and Minimum Points?

**cloud5** · July 26th 2009, 02:18 AM

1. $y=(1-x^2)(4-x^2)$

2. $y=3x^4-16x^3+18x^2$

Answer:
1. $(-\sqrt{\frac{5}{2}}, -2) (\sqrt{\frac{5}{2}}, -2)$

2. $(1, 5) (3, -27)$

**songoku** · July 26th 2009, 02:41 AM

The condition is : dy/dx = 0

but i get three values for each question, not two

**I-Think** · July 26th 2009, 02:49 AM

The only way I know to solve this question is to employ some calculus and differentiate. Differentiate to find $\frac{dy}{dx}$ , then set $\frac{dy}{dx}=0$ , and solve the resulting equation.

Eg. No.2

$\frac{dy}{dx}=12x^3-48x^2+36x$
$=12x(x^2-4x+3)$
$\frac{dy}{dx}=12x(x^2-4x+3)=0$
$=(12x)(x-1)(x-3)=0$
$x=0,1,3$

Place these results into the original equation

When $x=1$
$y=3(1)^4-16(1)^3+18(1)^2=5$
When $x=3$
$y=3(3)^4-16(3)^3+18(3)^2=-27$
When $x=0,y=0$

So your turning points are

$(0,0)$

Do the same for no.1

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