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

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Answer 1 · 2017-01-02T01:39:36

$h (x)$ $h(x)$ is the standard function $ln (x)$ $ln(x)$ shifted (transformed) one unit left (negative) on the $x$ $x$ -axis

$h (x) = ln (x + 1)$ $h(x) = ln(x+1)$

$ln (x)$ $ln(x)$ is defined for $x > 0 \to h (x)$ $x>0 -> h(x)$ is defined for $x + 1 > 0$ $x+1>0$
$:. h(x)$ is defined for $x> -1$

$ln(1) = 0 -> h(x) = 0$ for $x+1 = 1$
$:. h(x) = 0$ for $x=0$

$h(x)$ is the standard function $ln(x)$ shifted (transformed) one unit left (negative) on the $x$ -axis

The graph of $h(x)$ is shown below:

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

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