Find Minima and Maxima of $ y = \frac{x^2-3x+2}{x^2+2x+1}$

Question

$$ y = \frac{x^2-3x+2}{x^2+2x+1}$$

I guess I made some mistakes cause after taking the first derivative and simlifying I have

$$y = \frac{2x^3-4x^2+5}{(x+1)^2}$$

but then numerator has complex roots. which should not be, IMO

Answer 1

Well, you can use five rules:

1. Quotient rule: $$\frac{\text{d}}{\text{d}x}\left(\frac{\text{u}}{\text{v}}\right)=\frac{\displaystyle\text{v}\cdot\frac{\text{d}\text{u}}{\text{d}x}-\text{u}\cdot\frac{\text{d}\text{v}}{\text{d}x}}{\displaystyle\text{v}^2}\tag1$$
2. Differentiate the sum term by term and factor out constants.
3. The power rule: $$\frac{\text{d}}{\text{d}x}\left(x^\text{n}\right)=\text{n}x^{\text{n}-1}\tag2$$
4. The derivative of $x$ is $1$.
5. The derivative of $1$ is $0$:

So, using the quotient rule:

$$\text{y}'\left(x\right)=\frac{\displaystyle\left(x^2+2x+1\right)\cdot\frac{\text{d}}{\text{d}x}\left(x^2-3x+2\right)-\left(x^2-3x+2\right)\cdot\frac{\text{d}}{\text{d}x}\left(x^2+2x+1\right)}{\displaystyle\left(x^2+2x+1\right)^2}\tag3$$

Now, we can use the second rule to get:

• \begin{equation} \begin{split} \frac{\text{d}}{\text{d}x}\left(x^2-3x+2\right)&=\frac{\text{d}}{\text{d}x}\left(x^2\right)-3\cdot\frac{\text{d}}{\text{d}x}\left(x\right)+2\cdot\frac{\text{d}}{\text{d}x}\left(1\right)\\ \\ &=2x^{2-1}-3\cdot1+2\cdot0\\ \\ &=2x-3 \end{split}\tag4 \end{equation}
• \begin{equation} \begin{split} \frac{\text{d}}{\text{d}x}\left(x^2+2x+1\right)&=\frac{\text{d}}{\text{d}x}\left(x^2\right)+2\cdot\frac{\text{d}}{\text{d}x}\left(x\right)+\frac{\text{d}}{\text{d}x}\left(1\right)\\ \\ &=2x^{2-1}+2\cdot1+0\\ \\ &=2x+2 \end{split}\tag5 \end{equation}

So, we end up with:

\begin{equation} \begin{split} \text{y}'\left(x\right)&=\frac{\displaystyle\left(x^2+2x+1\right)\cdot\left(2x-3\right)-\left(x^2-3x+2\right)\cdot\left(2x+2\right)}{\displaystyle\left(x^2+2x+1\right)^2}\\ \\ &=\frac{5x-7}{\left(1+x\right)^3} \end{split}\tag6 \end{equation}

So, we get:

$$\frac{5x-7}{\left(1+x\right)^3}=0\space\Longleftrightarrow\space x=\frac{7}{5}\tag7$$

Answer 2

Hint.  Write it as:

$$ y = \frac{x^2+2x+1-5(x+1)+6}{(x+1)^2}=1 -\frac{5}{x+1}+\frac{6}{(x+1)^2}$$

The latter is a quadratic in $w=\frac{1}{x+1}\,$ with easy to determine extrema: $\;y=1-5w+6w^2\,$.

Answer 3

Let's derive

$$y'={(2x-3)(x^2+2x+1)-(2x+2)(x^2-3x+2)\over (x^2+2x+1)^2}={5x^2-2x-7\over (x+1)^4}={5x-7\over (x+1)^3}$$

Answer 4

The derivative is not correct, unfortunately.

Given that $x^2+2x+1 = (x+1)^2$, observe that if we let $g(x) = (x+1)^2$ and $f(x) = x^2-3x+2$ we have $$\begin{align}{dy\over dx} &= {g(x)f'(x) - f(x)g'(x) \over g^2(x)}\\ &= {(x+1)^2(2x-3) - (x^2-3x+2)2(x+1) \over (x+1)^4}\\ &= {(x+1)(2x-3)-2(x^2-3x+2) \over(x+1)^3}\\&= {2x^2-3x+2x-3 -2x^2+6x-4\over(x+1)^3}\\ &= {5x - 7\over(x+1)^3}. \end{align}$$

Now we set the numerator and the denominator equal to $0$ to find the points of the function where we have global maxima and global minima.

Answer 5

It is immediate to see that there is some mistake in the derivative. In the numerator we must have a term $2x^3$ by the product of the derivative of the numerator by the denominator and an opposite term $-2x^3$ from the product of the numerator by the derivative of the denominator. So we cannot have a third degree term.

Answer 6

$$y = \frac{x^2 - 3x + 2}{x^2 + 2x + 1} = 1 - \frac{5x-1}{(x+1)^2}$$ \begin{align*} \frac{dy}{dx} &= \frac{d}{dx} \left[ \frac{1-5x}{(x+1)^2} \right] \\ &= \frac{-5\cdot(x+1)^2-(1-5x)\cdot 2(x+1)}{(x+1)^4} \\ &= \frac{-5\cdot(x+1)-2(1-5x)}{(x+1)^3} \\ &= \frac{-5x-5-2+10x}{(x+1)^3} \\ &= \frac{5x-7}{(x+1)^3} \end{align*}

$$\frac{dy}{dx} = 0 \Longleftrightarrow x=\frac{7}{5} $$

Answer 7

Given, $y = \frac{x^2 - 3x + 2}{x^2 + 2x + 1} => yx^2 + 2xy + y = x^2 - 3x + 2 => (y - 1)x^2 + (2y + 3)x + y - 2 = 0$

Now since $x$ is needed to be real,
$b^2 - 4ac ≥ 0 => (2y + 3)^2 - 4(y - 1)(y - 2) ≥ 0 => y ≥ \frac{-1}{24}$