Question

select the correct intervals of increase and decrease for the function $f(x)=\log_4(x + 1)-3$

increase: $(-\infty,\infty)$

increase: $(-1,\infty)$

decrease: $(-3,\infty)$

increase: none

decrease: none

Explanation:

Step1: Determine the domain of the function

For the function \(y = \log_{a}u\) (\(a>0,a
eq1\)), the argument \(u=x + 1>0\), so \(x>-1\). The domain of \(f(x)=\log_{4}(x + 1)-3\) is \((-1,\infty)\).

Step2: Analyze the monotonicity of the logarithmic function

The general form of a logarithmic function is \(y=\log_{a}x\). When \(a > 1\) (in our case \(a = 4>1\)), the function \(y=\log_{a}x\) is an increasing function on its domain \((0,\infty)\).
For the function \(y=\log_{4}(x + 1)-3\), let \(u=x + 1\). The function \(y=\log_{4}u-3\) is a composition of \(y=\log_{4}u\) and \(u=x + 1\). The function \(u=x + 1\) is a linear function with a slope \(m = 1>0\) (increasing on \((-1,\infty)\)), and \(y=\log_{4}u\) is increasing on \((0,\infty)\).
By the chain - rule for composite functions (if \(y = f(g(x))\), and \(f\) and \(g\) are both increasing or both decreasing, then \(y=f(g(x))\) is increasing; if one is increasing and the other is decreasing, then \(y = f(g(x))\) is decreasing), since both \(y=\log_{4}u\) (with \(u=x + 1\)) and \(u=x + 1\) are increasing functions on the domain of \(f(x)\) (\(x>-1\)), the function \(y=\log_{4}(x + 1)-3\) is increasing on its domain \((-1,\infty)\).

Answer:

Increase: \((-1,\infty)\)