Help on this stupid ridiculously hard maths question?

Question

by means of sketch or otherwise determine the range of values of x for which

x(2x-3)(x-4) are less than zero

Answer 1

Since this is already factored, it will be easy to determine the zeroes of the function. It is when x = 0, x = 3/2 and x = 4

The function will be continuous between these points, so should check a point in between each to figure out whether it is positive or negative.

You have 3 points (0, 1.5, 4)

If you check a value of x < 0 (say -1), the result comes out negative.
If you check a value between 0 and 1.5 (say 1), the result comes out positive.
If you check a value between 1.5 and 3 (say 2), the result comes out negative.
If you check a value greater than 4 (say 5), the result comes out positive.

So the values of x where the function is negative:
x < 0 and 1.5 < x < 4

If it helps, here's a graph of the function:

Answer 2

A sketch would be easier, but I can't draw here, sorry.

It's not that bad. The zeros can be obtained by setting each of the multiplicative terms to zero:

x = 0
2x-3 = 0
x-4=0

So the function is 0 when x = 0, 1.5, or 4.

Negative values of x yield negative values of y, so y is < 0 for all x < 0. Since there is another zero at 1.5, is the bump between x=0 and x=1.5 positive or negative? Probably positive, but we can evaluate at x=1 and check: 1(-1)(-3)=3, OK, it's positive. The lump between x=1.5 and x=4 is probably negative, let's check x=2: 2(1)(-2)=2, so that lump is negative. finally how about x>4?: 5(7)(1)=35.

So after all that, x(2x-3)(x-4) is less than zero if x<0 or if 1.5 < x < 4

Answer 3

You want the product of these three factors to be negative (less than zero). The product of three factors is negative when all three factors are negative or just one of the factors is negative. If two are negative and one positive then the product is positive.

Solve and graph the following:

x < 0, 2x-3 < 0, and x-4 < 0 to get

x < 0, x < 3/2 , and x < 4

From these results or their graph you can see that all three factors are negative when x < 0, and exactl one of them is negative when 3/2 < x < 4 so the product will be negative when

x < 0 or 3/2 < x < 4 which can also be written as

(-infinity, 0) U (3/2, 4)

Answer 4

say f(x)=0
let f(x)=x(2x-3)(x-4)
so x(2x-3)(x-4)=0
x=0 or 2x-3=0 or x-4=0
x=0 or x=3/2 or x=4

put values into table. Try values within each of these inequalities and note whether it is more than or less than 0
x<0 f(x)<0
00
3/2 x>4 f(x)>0
Therefore x(2x-3)(x-4)<0 when x<0 or 3/2
This isn't that hard really... you should try hyperbolics (then again maybe you shouldn't!) anyway... hope that helped!

Answer 5

how you would do this problem is by finding the zeroes. to do that, you would set each equation equal to zero like this:
x=0
2x-3=0
x-4=0
then solve each equation for x and get x=0,3/2,4
now you know where the graph will be equal to zero. now make intervals.
x<0, 04
choose a value in each interval and plug it into the equation to see if it is less than zero. for example, you could choose -1, 1, 2, and 5. hope that helped!

Answer 6

x(2x-3)(x-4) = 0
a*b*c= 0 means either a, b or c is 0, so
x=0 or 2x-3 = 0 or x - 4 = 0
x=0 or x=1.5 or x=4

A sketch would be like this:
*edit, sketch didn't work out*
It looks a bit like an N, with the x-axis going through the middle.

So
if x < 0 it's less than zero and
if 1.5 < x < 4 it's less than zero

Answer 7

first find the zeros of the expression

x(2x-3)(x-4) =0
x =0 ; x = 3/2 ; x =4

Range where the expression is less than zero is = ] -infinity, 0[ and ]3/2 , 4 [

Answer 8

f(x)=0 at 0 , 3/2 & 4
so
it is -ve ]-∞,0[ & ]3/2 ,4 [

Answer 9

You just have to distribute

x(2x-3)(x-4)<0
2x-3x(x-4)<0
-x(x-4)<0
-x+4x<0
3x<0
x<0

So, x has to be less than zero

Answer 10

yup that is hard but if I give you the answer you will not learn anything ask your teacher for help.