How do you graph $y = 2 ln (x + 1)$ $y=2ln(x+1)$ ?

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Answer 1 · 2016-11-21T08:28:25

y is defined for $x > - 1$ $x> -1$ and has a single zero at $x = 0$ $x=0$

$y = 2 ln (x + 1)$ $y=2ln(x+1)$

$ln (x)$ $ln(x)$ is defined for $x > 0$ $x>0$
Hence $y$ $y$ is defined for $x > - 1$ $x> -1$

$ln (1) = 0$ $ln(1) =0$
Hence $ln (x + 1) = 0$ $ln(x+1) = 0$ for $(x + 1) = 1$ $(x+1) = 1$
$:. y$ has a single zero at $x=0$

The graph of $y$ is shown below:

graph{2ln(x+1) [-10, 10, -5, 5]}

How do you graph y=2ln(x+1)y=2ln(x+1) y=2ln(x+1)?