show that the function ( f(t)=sqrt{t}+sqrt{1 + t}-4 ) has exactly one zero in the interval ( (0,infty) )…

Explanation:

Step1: Check the continuity and values at endpoints

The function (f(t)=\sqrt{t}+\sqrt{1 + t}-4) is continuous for (t\in(0,\infty)) since (\sqrt{t}) and (\sqrt{1 + t}) are continuous for (t\geq0). When (t = 0), (f(0)=\sqrt{0}+\sqrt{1+0}-4=1 - 4=-3<0). As (t\rightarrow\infty), (\lim_{t\rightarrow\infty}f(t)=\lim_{t\rightarrow\infty}(\sqrt{t}+\sqrt{1 + t}-4)=\infty). By the Intermediate Value Theorem, there exists at least one (c\in(0,\infty)) such that (f(c) = 0).

Step2: Check the monotonicity

Find the derivative (f^\prime(t)) using the formula ((\sqrt{u})^\prime=\frac{u^\prime}{2\sqrt{u}}). If (u = t), then ((\sqrt{t})^\prime=\frac{1}{2\sqrt{t}}), and if (u=1 + t), then ((\sqrt{1 + t})^\prime=\frac{1}{2\sqrt{1 + t}}). So (f^\prime(t)=\frac{1}{2\sqrt{t}}+\frac{1}{2\sqrt{1 + t}}>0) for all (t\in(0,\infty)). A function with a positive - derivative is strictly increasing.

Answer:

Since (f(t)) is continuous on ((0,\infty)), (f(0)<0), (\lim_{t\rightarrow\infty}f(t)=\infty) (so there is at least one zero by the Intermediate Value Theorem) and (f(t)) is strictly increasing ( (f^\prime(t)>0) for (t\in(0,\infty))), the function (f(t)=\sqrt{t}+\sqrt{1 + t}-4) has exactly one zero in the interval ((0,\infty)).