How to find intervals where this function is decreasing and increasing?

[Mohmmad Maaitah]

Homework Statement

As in picture

Relevant Equations

I tried to solve it after first derivative and I get x=0 and a some other equation that when I use the quad formula I get minus in the root so what do I do to solve it

Please walk me step by step on how to do it (we don't have imaginary numbers so don't bring that up)
Also how to put signs on the numbers line when I get minus in the root? (non solveable equation)
Sorry for my English.

 

[fresh_42]

There is a method that most often works:

a) Find all derivatives, here $f(x) \, , \, f'(x)\, , \,f''(x)\, , \,f'''(x).$
b) Determine all zeros of these derivatives, including $f(x).$
c) Determine $\displaystyle{\lim_{n \to \infty}f(n)}$ and $\displaystyle{\lim_{n \to -\infty}} f(n).$

With that information you can almost draw the entire function: you have the intersections with the $x$-axis, the inclination points, minima and maxima, and what it does at the left and right of your paper.

Only inclination points might be a bit tricky. You can have $f'(x)=0$ without being a minimum or a maximum. $f''(x)\lessgtr 0$ defines a right- or left curve. (I always think of the standard parabola $y=x^2$ to remember which is which).

 

[Mohmmad Maaitah]

But how to draw the increase and decrease using f'(x) alone?
How to factor what I get
I cant get that

 

[fresh_42]

But how to draw the increase and decrease using f'(x) alone?

That is the fourth step. Start with step one, two and three!

We start with step one - differentiation:
\begin{align*}
f(x)&=x^4-5x^3+9x^2=x^2(x^2-5x+9)\\
f'(x)&=4x^3-15x^2+18x=x(4x^2-15x+18)\\
f''(x)&=12x^2-30x+18=6(x-1)(2x-3)
\end{align*}

Step two - zeros:
I do not know which method you use to determine whether a quadratic function has real solutions or not. In this case, both quadratic terms are positive:
\begin{align*}
x^2-5x+9 &> 0\text{ for all }x\in \mathbb{R}\\
4x^2-15x+18&>0\text{ for all }x\in \mathbb{R}
\end{align*}

This means that everything of interest happens at $x=0$ as the only zero of $f(x)$ and $f'(x)$. However, there is a change in curvature, between right to left curve or the other way around. We will see later. For now,
$$
f''(x)=0 \Longrightarrow x=1 \text{ or }x=1.5
$$

But we still have step three - asymptotic behaviour:
\begin{align*}
\displaystyle{\lim_{n \to \infty}}f(n)&= +\infty \\
\displaystyle{\lim_{n \to -\infty}}f(n)&=+\infty
\end{align*}

This means, so far we have

and something interesting going on at $x=1$ and $x=1.5.$

How to factor what I get
I cant get that

Now, investigate what happens at $f(0)=0.$ How does the function behave left and right of it?

If you have done this, then we consider what happens to the curvature, which is $f''(x),$ to the left of $x=1,$ between $1<x<1.5$ and to the right of $x=1.5.$

 

[Mohmmad Maaitah]

That is the fourth step. Start with step one, two and three!

We start with step one - differentiation:
\begin{align*}
f(x)&=x^4-5x^3+9x^2=x^2(x^2-5x+9)\\
f'(x)&=4x^3-15x^2+18x=x(4x^2-15x+18)\\
f''(x)&=12x^2-30x+18=6(x-1)(2x-3)
\end{align*}

Step two - zeros:
I do not know which method you use to determine whether a quadratic function has real solutions or not. In this case, both quadratic terms are positive:
\begin{align*}
x^2-5x+9 &> 0\text{ for all }x\in \mathbb{R}\\
4x^2-15x+18&>0\text{ for all }x\in \mathbb{R}
\end{align*}

This means that everything of interest happens at $x=0$ as the only zero of $f(x)$ and $f'(x)$. However, there is a change in curvature, between right to left curve or the other way around. We will see later. For now,
$$
f''(x)=0 \Longrightarrow x=1 \text{ or }x=1.5
$$

But we still have step three - asymptotic behaviour:
\begin{align*}
\displaystyle{\lim_{n \to \infty}}f(n)&= +\infty \\
\displaystyle{\lim_{n \to -\infty}}f(n)&=+\infty
\end{align*}

This means, so far we have

View attachment 326451

and something interesting going on at $x=1$ and $x=1.5.$
Now, investigate what happens at $f(0)=0.$ How does the function behave left and right of it?

If you have done this, then we consider what happens to the curvature, which is $f''(x),$ to the left of $x=1,$ between $1<x<1.5$ and to the right of $x=1.5.$

This is so much help thank you very much!

 

[fresh_42]

This is so much help thank you very much!

You are not done yet. There is still to decide whether $f(0)$ is a touching point or an intersection, and what happens at the part between, say $x\in (0.5\, ; \,2).$

 

[Mohmmad Maaitah]

You are not done yet. There is still to decide whether $f(0)$ is a touching point or an intersection, and what happens at the part between, say $x\in (0.5\, ; \,2).$

My main struggle was not noting
\begin{align*}

x^2-5x+9 &> 0\text{ for all }x\in \mathbb{R}\\

4x^2-15x+18&>0\text{ for all }x\in \mathbb{R}

\end{align*}

 

[fresh_42]

My main struggle was not noting
\begin{align*}

x^2-5x+9 &> 0\text{ for all }x\in \mathbb{R}\\

4x^2-15x+18&>0\text{ for all }x\in \mathbb{R}

\end{align*}

Which procedures do you know?

You could do it by a formula (we called it p-q formula, but I have also seen versions with a,b,c), by completing the square, calculating the discriminant, or by Viète's equation system. Of course, they are all the same, just in different wordings, but one of them should be familiar to you.

 

[Mohmmad Maaitah]

I can't factor any of those problems
(so I can find increase and decrease intervals)

 

[Mohmmad Maaitah]

I did factor a lot of other problems but factoring the first derivatives of these is hard to me

 

[fresh_42]

A method that should work without memorizing formulas is the completion of the square. Say we have $y=x^2-5x+9.$ Then $$y=x^2-5x+9=x^2+2\cdot \dfrac{5}{2}\cdot x +9=\underbrace{x^2-2\cdot \dfrac{5}{2}\cdot x+ \left(\dfrac{5}{2}\right)^2}_{=(x-2.5)^2\;\geq\; 0}+\underbrace{\left(9-\left(\dfrac{5}{2}\right)^2\right)}_{=:D}$$

Finally, $D>0$ so the entire expression is greater than zero. In a formula for the zeros, we would get the square root of $-D$ which has no real solutions.