find min 46

Note that the function $f(x) = x + \sqrt{x+1} + 5$ is a monotonic increasing function. That is the larger the value of $x$ , the larger the value of $f(x)$ . And the smallest value of $x$ possible is $x = -1$ , because for $x < -1$ , $f(x)$ is unreal. Therefore $\min(f(x)) = f(\min(x)) = f(-1) = -1 + \sqrt 0 + 5 = \boxed 4$ .

$x+\sqrt {x+1}+5=\left (\sqrt {x+1}+\dfrac 12\right )^2+\dfrac {15}{4}$

Since $\sqrt {x+1}\geq 0$ , therefore the minimum of the expression is $\dfrac 14+\dfrac {15}{4}=\boxed 4$ .

Upon observation, $\frac{dy}{dx} = 1 + \frac{1}{2\sqrt{x+1}}$ which is strictly positive over $(-1, \infty).$ This implies $y(x)$ is a strictly monotonically increasing function over its domain $x \in [-1, \infty)$ , and its minimum value is attained at $x = -1 \Rightarrow \boxed{y(-1) = 4}.$