For what values of x is dy/dx zero and undefined?

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For what values of x is dy/dx zero and undefined?

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

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Answer 1 · 2018-02-22T11:10:39

$\frac{d y}{d x}$ $dy/dx$ is zero for $x = - 2 \pm \sqrt{11}$ $x =-2 pm sqrt(11)$ , and $\frac{d y}{d x}$ $dy/dx$ is undefined for $x = - 2$ $x=-2$

Explanation:

Find the derivative:
$\frac{d y}{d x} = \frac{d (x^{2} - 3 x + 1)}{d x} \frac{1}{x + 2} + (x^{2} - 3 x + 1) \frac{d}{d x} (\frac{1}{x + 2})$ $dy/dx = (d(x^2 - 3x + 1))/dx 1/(x+2) + (x^2 - 3x + 1)(d)/(dx)(1/(x + 2))$
$= \frac{2 x - 3}{x + 2} - (x^{2} - 3 x + 1) \frac{1}{{(x + 2)}^{2}}$ $=(2x-3)/(x+2) - (x^2 - 3x + 1)1/(x + 2)^2$
$= \frac{(2 x - 3) (x + 2) - (x^{2} - 3 x + 1)}{{(x + 2)}^{2}}$ $=((2x-3)(x+2) - (x^2 - 3x + 1))/(x + 2)^2$
$= \frac{2 x^{2} - 3 x + 4 x - 6 - x^{2} + 3 x - 1}{{(x + 2)}^{2}}$ $= (2x^2 - 3x + 4x -6 - x^2+3x-1)/(x + 2)^2$
$= \frac{x^{2} + 4 x - 7}{{(x + 2)}^{2}}$ $= (x^2 + 4x -7)/(x + 2)^2$
by the product rule and various simplifications.

Find zeros:
$\frac{d y}{d x} = 0$ $dy/dx=0$ if and only if $x^{2} + 4 x - 7 = 0$ $x^2 + 4x -7=0$ .
The roots of this polynomial are
$x_{1, 2} = (\frac{1}{2}) (- 4 \pm \sqrt{4^{2} - 4 (- 7)}) = - 2 \pm \sqrt{11}$ $x_{1,2} = (1/2)(-4 pm sqrt(4^2 - 4(-7))) = -2 pm sqrt(11)$ ,
so $\frac{d y}{d x} = 0$ $dy/dx=0$ for $x = - 2 \pm \sqrt{11}$ $x = -2 pm sqrt(11)$ .

Find where $\frac{d y}{d x}$ $dy/dx$ is undefined:
Since division by $0$ $0$ is not allowed, $\frac{d y}{d x}$ $dy/dx$ is undefined where ${(x + 2)}^{2} = 0$ $(x + 2)^2=0$ , that is, where
$x = - 2$ $x=-2$ .

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For what values of x is dy/dx zero and undefined?

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

1 Answer

Explanation:

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