What is the f(x)=sqrt(3x-2)+(2x)/(x^2-4) ?

The f(x)=sqrt(3x-2)+(2x)/(x^2-4) is

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f(x)=sqrt(3x-2)+(2x)/(x^2-4)

f (x) = 3 x - 2 + \frac{2 x}{x ^{2} - 4}

Domain : \frac{2}{3} \leq x < 2 or x > 2

Range : f (x) \leq 0.35170 \dots or f (x) \geq 3.76269 \dots

X Intercepts : (0.72465 \dots, 0), (1.45191 \dots, 0)

Asymptotes : Vertical x = 2

Extreme Points : Minimum (\frac{2}{3}, - \frac{3}{8}), Maximum (1.09449 \dots, 0.35170 \dots), Minimum (3.43327 \dots, 3.76269 \dots)

Interval Notation

Domain : [\frac{2}{3}, 2) \cup (2, \infty)

Range : (- \infty, 0.35170 \dots] \cup [3.76269 \dots, \infty)

Solution steps

Domain of

3 x - 2 + \frac{2 x}{x ^{2} - 4} : \frac{2}{3} \leq x < 2 or x > 2

Range of

3 x - 2 + \frac{2 x}{x ^{2} - 4} : f (x) \leq 0.35170 \dots or f (x) \geq 3.76269 \dots

Axis interception points of

3 x - 2 + \frac{2 x}{x ^{2} - 4} :

X Intercepts

: (0.72465 \dots, 0), (1.45191 \dots, 0)

Asymptotes of

3 x - 2 + \frac{2 x}{x ^{2} - 4} :

Vertical

: x = 2

Extreme Points of

3 x - 2 + \frac{2 x}{x ^{2} - 4} :

Minimum

(\frac{2}{3}, - \frac{3}{8}),

Maximum

(1.09449 \dots, 0.35170 \dots),

Minimum

(3.43327 \dots, 3.76269 \dots)