Find all critical points of the function $\newline$ $f(x)=\frac{3x^{2}}{7x^{2}-2x+2}.$ $\newline$ (Use symbolic notation and fractions where needed. Give your answer in the form of a comma separated list. If the function does not have any critical points, enter DNE.) $\newline$ critical points:

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Q. Find all critical points of the function

\newline

f(x)=\frac{3x^{2}}{7x^{2}-2x+2}.

\newline

(Use symbolic notation and fractions where needed. Give your answer in the form of a comma separated list. If the function does not have any critical points, enter DNE.)

\newline

critical points:

Calculate Derivative: To find the critical points of the function $f(x) = \frac{3x^2}{7x^2 - 2x + 2}$ , we need to find the values of $x$ where the derivative $f'(x)$ is zero or undefined.
Apply Quotient Rule: First, we calculate the derivative of $f(x)$ using the quotient rule, which states that if $f(x) = \frac{g(x)}{h(x)}$ , then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$ . Here, $g(x) = 3x^2$ and $h(x) = 7x^2 - 2x + 2$ .
Simplify Numerator: Compute the derivative of $g(x) = 3x^2$ , which is $g'(x) = 6x$ .
Find Critical Points: Compute the derivative of $h(x) = 7x^2 - 2x + 2$ , which is $h'(x) = 14x - 2$ .
Factor Common Term: Apply the quotient rule to find $f'(x)$ : $f'(x) = \frac{(6x \cdot (7x^2 - 2x + 2) - 3x^2 \cdot (14x - 2))}{(7x^2 - 2x + 2)^2}.$
Solve for x: Simplify the numerator of $f'(x)$ : $\text{Numerator} = 6x \cdot (7x^2 - 2x + 2) - 3x^2 \cdot (14x - 2) = 42x^3 - 12x^2 + 12x - 42x^3 + 6x^2 = -12x^2 + 12x + 6x^2 = -6x^2 + 12x.$
Check Denominator: Set the simplified numerator equal to zero to find the critical points: $\newline$ $-6x^2 + 12x = 0$ .
Identify Critical Points: Factor out the common term: $\newline$ $-6x(x - 2) = 0$ .
Identify Critical Points: Factor out the common term: $\newline$ $-6x(x - 2) = 0$ .Set each factor equal to zero and solve for $x$ : $\newline$ $-6x = 0$ gives $x = 0$ . $\newline$ $x - 2 = 0$ gives $x = 2$ .
Identify Critical Points: Factor out the common term: $\newline$ $-6x(x - 2) = 0$ . Set each factor equal to zero and solve for $x$ : $\newline$ $-6x = 0$ gives $x = 0$ . $\newline$ $x - 2 = 0$ gives $x = 2$ . Check if the denominator of $f'(x)$ becomes zero at $x = 0$ or $x = 2$ , which would make the derivative undefined: $\newline$ $(7x^2 - 2x + 2)^2$ at $x = 0$ is $x$ $1$ , which is not zero. $\newline$ $(7x^2 - 2x + 2)^2$ at $x = 2$ is $x$ $4$ , which is $x$ $5$ , which is not zero.
Identify Critical Points: Factor out the common term: $\newline$ $-6x(x - 2) = 0$ . Set each factor equal to zero and solve for $x$ : $\newline$ $-6x = 0$ gives $x = 0$ . $\newline$ $x - 2 = 0$ gives $x = 2$ . Check if the denominator of $f'(x)$ becomes zero at $x = 0$ or $x = 2$ , which would make the derivative undefined: $\newline$ $(7x^2 - 2x + 2)^2$ at $x = 0$ is $x$ $1$ , which is not zero. $\newline$ $(7x^2 - 2x + 2)^2$ at $x = 2$ is $x$ $4$ , which is $x$ $5$ , which is not zero. Since the denominator of $f'(x)$ does not become zero at $x = 0$ or $x = 2$ , the critical points are $x = 0$ and $x = 2$ .