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Stage 4 - Stage 5

Lesson

There are algebraic and graphical methods that we can use to solve inequalities involving the cubes of unknowns. We'll discuss a couple of examples.

Solve $\left(x-2\right)^3\le-27$(`x`−2)3≤−27.

While this is a straightforward example, try to pay close attention to the strategies involved. The same strategies can be used for much more difficult questions.

Using an algebra approach, we write:

$\left(x-2\right)^3$(x−2)3 |
$<=$<= | $-27$−27 |

$x-2$x−2 |
$<=$<= | $-3$−3 |

$x$x |
$<=$<= | $-1$−1 |

Using a graphical approach we can graph $f\left(x\right)=\left(x-2\right)^3$`f`(`x`)=(`x`−2)3 and $g\left(x\right)=-27$`g`(`x`)=−27 on the same set of axes, and then see for what values of $x$`x` is $f\left(x\right)\le g\left(x\right)$`f`(`x`)≤`g`(`x`). The result, shown graphically is $x\le-1$`x`≤−1.

An alternative approach is to rearrange the inequality so that $\left(x-2\right)^3+27\le0$(`x`−2)3+27≤0, and form the function $h\left(x\right)=\left(x-2\right)^3+27$`h`(`x`)=(`x`−2)3+27. Then see for what $x$`x` is $h\left(x\right)\le0$`h`(`x`)≤0. Again, shown graphically, the same result is found.

Solve $2x^3-5x^2\ge28x-15$2`x`3−5`x`2≥28`x`−15

Using algebra, we can proceed as follows. You will note that a factorisation occurs in the steps. If you are unsure of how to factorise cubics, please refer to cubic factorisation.

$2x^3-5x^2$2x3−5x2 |
$>=$>= | $28x-15$28x−15 |

$2x^3-5x^2-28x+15$2x3−5x2−28x+15 |
$>=$>= | $0$0 |

$\left(x+3\right)\left(2x-1\right)\left(x-5\right)$(x+3)(2x−1)(x−5) |
$>=$>= | $0$0 |

Now the last inequality is interesting to explore logically. It essentially tells us that, for a given value of $x$`x`, the product of three numbers $\left(x+3\right)$(`x`+3), $\left(2x-1\right)$(2`x`−1) and $\left(x-5\right)$(`x`−5) needs to be greater than or equal to zero.

Think of a number like $x=-3.1$`x`=−3.1. The product becomes $\left(-3.1+3\right)\left(2\times-3.1-1\right)\left(-3.1-5\right)=-5.832$(−3.1+3)(2×−3.1−1)(−3.1−5)=−5.832 and, being negative, fails to satisfy our inequality. Note that this specific negative product involved *three* negative numbers.

If we next choose, say $x=-2.9$`x`=−2.9, the product will involve *one* positive number and *two* negative numbers, and will thus satisfy the inequality.

The switch from negative to positive on the first factor $\left(x+3\right)$(`x`+3) meant that $x=3$`x`=3 became critical to the sign of the product. In fact we call $x=3$`x`=3 a critical point,

Similarly, for the second factor, $\left(2x-1\right)$(2`x`−1), the number $x=\frac{1}{2}$`x`=12 becomes critical as well. The sign of the $\left(2x-1\right)$(2`x`−1) will switch across $x=\frac{1}{2}$`x`=12 and thus effect the sign of the entire product.

The sign of the factor $\left(x-5\right)$(`x`−5) will also switch across $x=5$`x`=5.

Thus $x=-3$`x`=−3 , $x=\frac{1}{2}$`x`=12 and $x=5$`x`=5 are the three critical points.

The logic naturally gives rise to the following critical point diagram:

The diagram shows four intervals created by the three critical points. The product given by $\left(x+3\right)\left(2x-1\right)\left(x-5\right)$(`x`+3)(2`x`−1)(`x`−5) will be, from the left-most interval to the right-most, negative, positive, negative and positive respectively.

Hence, because our solution requires the product to be greater than or equal to zero, the solution is given by all $x$`x` such that $-3\le x\le\frac{1}{2}$−3≤`x`≤12 and $x\ge5$`x`≥5.

A simple graphical approach would be to first form the function given by $y=\left(x+3\right)\left(2x-1\right)\left(x-5\right)$`y`=(`x`+3)(2`x`−1)(`x`−5) (the fact that we have used the factored form is irrelevant - we could just as well use the form $y=2x^3-5x^2-28x+15$`y`=2`x`3−5`x`2−28`x`+15 ).

Then using technology, or otherwise, sketch the function and note for what intervals is it greater than or equal to zero. The graph below shows this. Note that there is an exact correspondence between the critical point diagram above and the behaviour of the function across the roots. The curve moves from negative to positive to negative and to positive as well.

An alternative approach would be to form two functions $f\left(x\right)=2x^3-5x^2$`f`(`x`)=2`x`3−5`x`2, and and $g\left(x\right)=28x-15$`g`(`x`)=28`x`−15 . Then graph both of these, and see where $f\left(x\right)\ge g\left(x\right)$`f`(`x`)≥`g`(`x`). The same solution will present itself.

Solve the inequality $\left(x+2\right)^3>-64$(`x`+2)3>−64.

Use the graph of the function $y=\left(x-1\right)\left(x+1\right)\left(x-3\right)$`y`=(`x`−1)(`x`+1)(`x`−3), provided below, to solve the inequality $\left(x-1\right)\left(x+1\right)\left(x-3\right)\le0$(`x`−1)(`x`+1)(`x`−3)≤0.

Loading Graph...

Select all correct regions.

$1\le x\le3$1≤

`x`≤3A$x\le-1$

`x`≤−1B$x\ge3$

`x`≥3C$-1\le x\le1$−1≤

`x`≤1D$x\ge3$

`x`≥3E

Use the graph of the function $y=2x^3-4x^2-30x$`y`=2`x`3−4`x`2−30`x` and the line $y=-72$`y`=−72 below to solve the inequality $2x^3-4x^2-30x<-72$2`x`3−4`x`2−30`x`<−72.

Loading Graph...

Solve the inequality $\left(x+2\right)^3>-64$(`x`+2)3>−64.

Use the graph of the function $y=\left(x-1\right)\left(x+1\right)\left(x-3\right)$`y`=(`x`−1)(`x`+1)(`x`−3), provided below, to solve the inequality $\left(x-1\right)\left(x+1\right)\left(x-3\right)\le0$(`x`−1)(`x`+1)(`x`−3)≤0.

Loading Graph...

Select all correct regions.

$1\le x\le3$1≤

`x`≤3A$x\le-1$

`x`≤−1B$x\ge3$

`x`≥3C$-1\le x\le1$−1≤

`x`≤1D$x\ge3$

`x`≥3E

Use the graph of the function $y=2x^3-4x^2-30x$`y`=2`x`3−4`x`2−30`x` and the line $y=-72$`y`=−72 below to solve the inequality $2x^3-4x^2-30x<-72$2`x`3−4`x`2−30`x`<−72.

Loading Graph...