okay so i have been given a question of :

on the graph of $\displaystyle y = 2x^3 -9x^2 - 12x - 7$

, 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.

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

3. Originally Posted by the_sensai
okay so i have been given a question of :

on the graph of $\displaystyle y = 2x^3 -9x^2 - 12x - 7$

, 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.
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.

4. thankyou so much that was a massive help.