# Algebra question?

## $f \left(x\right) = {x}^{3} - {x}^{2} - x - \left(\frac{11}{4}\right) .$ If $k$ is a constant such that the equation $f \left(x\right) = k$ has three real solutions, what could be the value of $k$?

Jul 20, 2018

$- 2.5648 > k > - 3.7500$ (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 $f \left(x\right)$ have more than 1 value for $x$
graph{x^3-x^2-x-2.75 [-10, 10, -5, 5]}

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

At the inflection points the slope of $f \left(x\right)$ = 0 so $d \left(f \left(x\right)\right) . \mathrm{dx} = 0$

$d \left(f \left(x\right)\right) . \mathrm{dx} = 3 {x}^{2} - 2 x - 1$

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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(3 \cdot - 1\right)}}{2 \cdot 3}$

$x = \frac{2 \pm \sqrt{4 + 12}}{6}$

$x = \frac{2 \pm \sqrt{16}}{6}$

$x = \frac{2}{6} \pm \frac{4}{6}$

$x = - \frac{1}{3} \mathmr{and} 1$ ...these are the values for $x$ at the inflection points

put theses into $f \left(x\right)$ to get the values asked for in the question

f(x)=x^3−x^2−x−2.75

at $x = \frac{1}{3}$
f(x)=(-1/3)^3−(-1/3)^2−(-1/3)−2.75

evaluate this now, or simplify the fractions:

$= - \frac{1}{27} - \frac{1}{9} + \frac{1}{3} - 2.75$
$= - \frac{1}{27} - \frac{3}{27} + \frac{9}{27} - 2.75$
$= \frac{5}{27} - 2.75$
$= - 2.5648$

or
f(x)=(1)^3−(1)^2−(1)−2.75

$= - 3.75$

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