Hi There,
I really need help on this please:
Let f(x) = x2(x squared)  3x2 (3x squared)  24x
(i) Find the coordinates of the local maximum point and the local minimum point on the curve f(x).
(ii) Show that the curve of f(x) intersects the xaxis at x = 0.
Thanks in advance. :)

kevin52
Where was this question taken from? Looks like a misprint. Traditionally you're asked to find the local max and min points of either a Cubic or a Quadratic function. The question you're asking is neither.

jamielovesmusic123
This question was taken from my textbook (Active Maths 3) Book 1

Richie_97
are you sure its 24x and not just 24?

Dazzla16
I am taking the assumption that your function f(x) = x^3  3x^2  24x
(i) First we find the derived function.
dy/dx = 3x^2  6x  24
Then let the derived function equal 0 (because the tangents of the local max and local min points are both 0)
3x^2  6x  24 = 0
x^2  2x  8 = 0 (dividing both sides by 3)
(x  4)(x + 2) = 0
x = 4 or x = 2
When x = 4,
y = (4)^3  3(4)^2  24(4) = 80
One of the local points is (4, 80)
When x = 2
y = (2)^3  3(2)^2  24(2) = 28
The other local point is (2, 28)
To determine which of these are local max and min points, we first have to find the second derivative (i.e. the derivative of the derivative of your original function).
d^2y/dx^2 = 6x  6
Then we input the x values of our local points into the second derivative. If the second derivative is negative, the point is a local max. If the second derivative is positive, the point is a local min.
At x = 4
6(4)  6 = 18
Since it's positive, the point (4, 80) is the local min.
At x = 2
6(2)  6 = 18
Since it's negative, the point (2, 28) is the local max.
If you're not thought this (the part where I determined which point is local min and which is local max) in OL, you can simply figure that out by looking at the yvalue of your two points. Since (2, 28) has a larger yvalue, it will be higher up the coordinate plane. Therefore it's the local max. Since (4, 80) has a smaller yvalue, it follows that it's the local min.
(ii) All you have to is input 0 into the function.
When x = 0, f(x) = (0)^3  3(0)^2  24(0) = 0
Since it passes through the point (0, 0) it intersects the xaxis at x = 0. Note that at any point on the xaxis, y = 0.
Hope this helps :)

Dazzla16
The start of (ii) is meant to say "All you have to do is input 0 into the function."

Dazzla16
@Richie_97 if you think about it, the function can't have a constant term (such as 24) if it's going to pass through the xaxis at x = 0 (part (ii) tells us to show this). The constant term acts as the y intercept, as inputting 0 for x will affect all terms (turning them into 0) except for the constant. Generally, for a function ax^3 + bx^2 + cx + d, it will pass through the point (0,d).

jamielovesmusic123
Thanks everyone for this, It actually really helped :)

Eoghan149
wat mathes

Carlos_8071
Hi

Eoghan149
What math

Me
Share files from your computer