I really need help on this please:
Let f(x) = x2(x squared) - 3x2 (3x squared) - 24x
(i) Find the co-ordinates 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 x-axis at x = 0.
Thanks in advance. :)
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.
are you sure its 24x and not just 24?
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 y-value of your two points. Since (-2, 28) has a larger y-value, it will be higher up the co-ordinate plane. Therefore it's the local max. Since (4, -80) has a smaller y-value, 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 x-axis at x = 0. Note that at any point on the x-axis, y = 0.
Hope this helps :)
The start of (ii) is meant to say "All you have to do is input 0 into the function."
@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 x-axis 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).