What is the f(x)=x^2+2ln(4/x)-e^{x-4} ?

The f(x)=x^2+2ln(4/x)-e^{x-4} is Domain: x>0 Intercepts: None Asymptotes: Vertical x=0

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f(x)=x^2+2ln(4/x)-e^{x-4}

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

Domain : x > 0

Intercepts : None

Asymptotes : Vertical x = 0

Interval Notation

Domain : (0, \infty)

Solution steps

Domain of

x^{2} + 2 ln (\frac{4}{x}) - e^{x - 4} : x > 0

Axis interception points of

x^{2} + 2 ln (\frac{4}{x}) - e^{x - 4} :

None

Asymptotes of

x^{2} + 2 ln (\frac{4}{x}) - e^{x - 4} :

Vertical

: x = 0

f (x) = lo g_{2} (x) + lo g_{4} (x)

y = \frac{x ^{2} - 4}{x ^{2} - 1}

f (x) = 3 x^{2} - 6 x + 10

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

f (x) = lo g_{6} (x - 1) - 5

What is the f(x)=x^2+2ln(4/x)-e^{x-4} ?
The f(x)=x^2+2ln(4/x)-e^{x-4} is Domain: x>0<br/>Intercepts: None<br/>Asymptotes: Vertical x=0