What is the f(x)= 3/(x-3)+(sqrt(4-x^2-1))/(1-|x-2|) ?

The f(x)= 3/(x-3)+(sqrt(4-x^2-1))/(1-|x-2|) is

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f(x)= 3/(x-3)+(sqrt(4-x^2-1))/(1-|x-2|)

f (x) = \frac{3}{x - 3} + \frac{4 - x ^{2} - 1}{1 - ∣ x - 2 ∣}

Domain : - 3 \leq x < 1 or 1 < x \leq 3

Range : - \infty < f (x) < \infty

X Intercepts : (1.46866 \dots, 0), Y Intercepts : (0, - 1 - 3)

Asymptotes : Vertical x = 1

Extreme Points : Maximum (- 3, - \frac{3 - 3}{2}), Minimum (3, - \frac{3 + 3}{2})

Solution steps

Domain of

\frac{3}{x - 3} + \frac{4 - x ^{2} - 1}{1 - ∣ x - 2 ∣} : - 3 \leq x < 1 or 1 < x \leq 3

Range of

\frac{3}{x - 3} + \frac{4 - x ^{2} - 1}{1 - ∣ x - 2 ∣} : - \infty < f (x) < \infty

Axis interception points of

\frac{3}{x - 3} + \frac{4 - x ^{2} - 1}{1 - ∣ x - 2 ∣} :

X Intercepts

: (1.46866 \dots, 0),

Y Intercepts

: (0, - 1 - 3)

Asymptotes of

\frac{3}{x - 3} + \frac{4 - x ^{2} - 1}{1 - ∣ x - 2 ∣} :

Vertical

: x = 1

Extreme Points of

\frac{3}{x - 3} + \frac{4 - x ^{2} - 1}{1 - ∣ x - 2 ∣} :

Maximum

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

Minimum

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

f (x) = 49 x^{2} - 144

f (n) = n^{2} - 15

f (x) = \frac{- ( x + 3 ) ( x - 1 )}{( x - 1 ) ( x + 2 )}

y = ln (x^{4} + 3 x^{2} + 1)

f (x) = \frac{2 + x}{x}

What is the f(x)= 3/(x-3)+(sqrt(4-x^2-1))/(1-|x-2|) ?
The f(x)= 3/(x-3)+(sqrt(4-x^2-1))/(1-|x-2|) is