Question

Algebra question?

`f(x)=x^3-x^2-x-(11/4).`
If `k` is a constant such that the equation `f(x)=k` has three real solutions, what could be the value of `k`?

Answer 1

`-2.5648>k> -3.7500` (4dp)

Explanation:

(I have included steps that could be skipped ...e.g. this can be solved quickly by graphing, or eqns can be evaluated in one step with a calculator.
Hopefully this helps in understanding each step, although it looks a bit long).

`f(x)=x^3−x^2−x−(11/4)`

`=x^3−x^2−x−2.75`

If we graph this function we can see where the values for `f(x)` have more than 1 value for `x`
graph{x^3-x^2-x-2.75 [-10, 10, -5, 5]}

If the value of `f(x)` is exactly at either of the 2 inflection point values there are 2 possible values for `X`, and between these two `f(x)` values there are 3 possible values for `X`.

At the inflection points the slope of `f(x)` = 0 so `d(f(x)).dx=0`

`d(f(x)).dx = 3x^2-2x-1`

so if we solve for `3x^2-2x-1 = 0` we will get these points.

Use a the quadratic formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(-2)+-sqrt((-2)^2-4(3*-1)))/(2*3)`

`x=(2+-sqrt(4+12))/(6)`

`x=(2+-sqrt(16))/6`

`x = 2/6+-4/6`

`x = -1/3 or 1` ...these are the values for `x` at the inflection points

put theses into `f(x)` to get the values asked for in the question

`f(x)=x^3−x^2−x−2.75`

at `x=1/3`
`f(x)=(-1/3)^3−(-1/3)^2−(-1/3)−2.75`

evaluate this now, or simplify the fractions:

`= -1/27 - 1/9 + 1/3 -2.75`
`= -1/27-3/27 + 9/27 - 2.75`
`= 5/27 - 2.75`
`= -2.5648`

or
`f(x)=(1)^3−(1)^2−(1)−2.75`

`= -3.75`

So `f(x)` has 3 real solutions when `f(x)` is < `-2.5648` and > `-3.75` (but only 2 at these values).