Graph of function intersects line

[hopelesss]

Explain why the graph of the function g (x) = x^3 - 2x + 1 intersects the line y = x in
each of the intervals (-2, -1), (0, 1) and (1, 2).

i hope someone can help me with this. To be honest i do not even know where to start.
i sketched g(x) and y=x and i don't see them intersect in the intervals... so i guess I am supposed to do something with g(x) first?

 

[SuperSonic4]

Explain why the graph of the function g (x) = x^3 - 2x + 1 intersects the line y = x in
each of the intervals (-2, -1), (0, 1) and (1, 2).

i hope someone can help me with this. To be honest i do not even know where to start.
i sketched g(x) and y=x and i don't see them intersect in the intervals... so i guess I am supposed to do something with g(x) first?

The question is asking you to find where the lines of g(x) = x and you can verify that it is true here: Wolfram|Alpha: Computational Knowledge Engine

The question is asking you to find the roots of the cubic (i.e. \(\displaystyle g(x)=0\) and compare them to the line y=x

Supposing the three roots (it's a cubic so has 3 roots) are \(\displaystyle a_1\), \(\displaystyle a_2\) and \(\displaystyle a_3\) then these points are \(\displaystyle (a_1,a_1) \ ,\ (a_2,a_2) \ ,\ (a_3,a_3)\)

Do you know how to (try) to factor a cubic? If you need a clue note that \(\displaystyle g(1)=0\)

Spoiler

so \(\displaystyle (x-1)\) is a factor of g(x) which can be written as \(\displaystyle g(x) = (x-1)(Ax^2+Bx+C)\) where A, B and C are constants

 

[MarkFL]

I would look at:

\(\displaystyle f(x)=g(x)-y=(x^3-2x+1)-(x)=x^3-3x+1\)

And then:

\(\displaystyle f'(x)=3x^2-3=3(x+1)(x-1)\)

We know then that $f$ has the following behavior:

Increasing on $(-\infty,-1)$

Decreasing on $(-1,1)$

Increasing on $(1,\infty)$

Now, we can compute:

\(\displaystyle f(-2)=-1\)

\(\displaystyle f(-1)=3\)

So, given the sign change of the function and its monotonic behavior on $(-2,-1)$, we know there is one root in that interval.

Can you complete the other two intervals?

 

[hopelesss]

thanks for responding to my post
yes i think i know how to factor a cubic.
i use polynomial factoring and get x^3-2x+1=(x-1)(x^2+x-1).
But i see MarkFL takes g(x)-y, then takes f'(x) before he factorise
and i did not do that so then its wrong. :)

I don't understand how i can know f has the behavior MarkFL gives

 

[MarkFL]

If a function, on some interval $(a,b)$, has a first derivative that does not change sign, then we know that function is strictly monotonic on that interval, that is, it either strictly increase or decreases. If we further find that the function has changed sign over that interval, then we know the function has 1 root somewhere in that interval. :D

 

[hopelesss]

so when i plot (0,1) and (1,2) into x^3-3x+1 i get
f(0)=1 and f(1)=-1
f(1)=-1 and f(2)=3

so i now know i have one root in each interval? what do this mean for my task?

 

[MarkFL]

so when i plot (0,1) and (1,2) into x^3-3x+1 i get
f(0)=1 and f(1)=-1
f(1)=-1 and f(2)=3

so i now know i have one root in each interval? what do this mean for my task?

This means you have completed the task. :)

 

[hopelesss]

I see i have much studying to do :)
thanks for all help, i will now know what to do if i get tasks like this again :)

 

[I like Serena]

Just because we can (and as an exercise): ;)

The graphs of $y=g(x)$ and $y=x$:
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[xmin=-3, xmax=3, ymin=-3, ymax=3]
\addplot[blue, thick, samples=101] (x, {x^3-2*x+1}) node at (-1.8,2) {$y=g(x)$};
\addplot[red, thick] (x, x) node at (-0.5,-1.2) {$y=x$};
\end{axis}
\end{tikzpicture}

The graph of $y=f(x)$ as defined by MarkFL.
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[xmin=-3, xmax=3, ymin=-3, ymax=3]
\addplot[blue, thick, samples=101] (x, {x^3-3*x+1}) node at (-2.2,2) {$y=f(x)$};
\addplot[red, thick] (x, 0) node at (-0.7,-0.3) {$y=0$};
\end{axis}
\end{tikzpicture}

And sign schemas of $f$ and $f'$, showing where the graph slopes up and down, and where we will find zeroes:

\begin{tikzpicture}

\draw (-3,2) node

{$f$} -- (3,2);
\draw (-2,1.9) node[below] {-2} -- (-2,2.1) node[above] {-1};
\draw (-1,1.9) node[below] {-1} -- (-1,2.1) node[above] {+3};
\draw ( 0,1.9) node[below] {0} -- ( 0,2.1) node[above] {+1};
\draw ( 1,1.9) node[below] {+1} -- ( 1,2.1) node[above] {-1};
\draw ( 2,1.9) node[below] {+2} -- ( 2,2.1) node[above] {+3};

\draw (-3,0) node

{$f'$} -- (3,0);
\node at (-2, 0.2) {$+$};
\draw (-1,-0.1) node[below] {-1} -- (-1,0.1) node[above] {0};
\node at (0, 0.2) {$-$};
\draw ( 1,-0.1) node[below] {+1} -- ( 1,0.1) node[above] {0};
\node at (2, 0.2) {$+$};

\end{tikzpicture}
We can see that we have 3 zeroes in respectively the intervals $(-2,-1), (0,1), (1,2)$.​