does y = -2 when x = 1 ???

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- Aug 20th 2008, 05:17 PM #1

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## help please

okay so i have been given a question of :

on the graph of

, the point (1, -2) is:

a minimum , a max , neither but on the graph or not on the graph.

So i have never done this before. Do i differentiate the equation and substitute values of x? Or is it a different way?

Thanks People your help is appreciated.

- Aug 20th 2008, 06:02 PM #2

- Aug 20th 2008, 06:28 PM #3

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If the graph is for y = 2x^3 -9x^2 -12x -7,

then point (1,-2) is not on the graph, because

-2 = 2(1^3) -9(1^2) -12(1) -7

-2 = -26 ----------not true.

But if the graph were for y = 2x^3 -9x^2**+**12x -7,

Then point (1,-2) is on the graph.

To check if (1,-2) is then a minimum, a maximum or neither, yes, you need to get the first derivative of y with respect to x. Then set dy/dx to zero.

y = 2x^3 -9x^2**+**12x -7

dy/dx = 6x^2 -18x +12

Set that to zero,

0 = 6x^2 -18x +12

Divide both sides by 6,

0 = x^2 -3x +2

0 = (x-1)(x-2)

x = 1 or 2

Meaning, when x=1 or x=2, the graph is minimum or maximum ....(or there is an inflection point).

When x=1,

y = 2x^3 -9x^2**+**12x -7

y = 2(1^3) -9(1^2) +12(1) -7

y = 2 -9 +12 -7 = -2

So, point (1,-2).

Do you know how to check if (1,-2) is minimum or maximum?

a) one way is find the dy/dx values when x is before and after x=1

b) another way is to find the concavity of the graph by getting the second derivative at x=1.

We do here the option b.

dy/dx = y' = 6x^2 -18x +12

d/dx(dy/dx) = y'' = 12x -18

At x=1,

y'' = 12(1) -18 = -6 ---------it is negative.

Meaning, the the graph at x=1 is concaved downward.

Meaning, the graph is a maximum there.

Therefore, (1,-2) is a maximum point.

- Aug 20th 2008, 06:33 PM #4

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