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

Explanation:

Step1: Check continuity and values at endpoints

The function (f(t)=\sqrt{t}+\sqrt{1 + t}-4) is continuous for (t>0) (sum of square - root functions which are continuous in their domains). 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)=\lim_{t\rightarrow\infty}\sqrt{t}(1+\sqrt{1+\frac{1}{t}})-4=\infty>0).

Step2: Apply the Intermediate Value Theorem

By the Intermediate Value Theorem (since (f(t)) is continuous on ((0,\infty)) and (f(0)<0), (f(t)\rightarrow\infty>0) as (t\rightarrow\infty)), there exists at least one (c\in(0,\infty)) such that (f(c) = 0).

Step3: Check the derivative for monotonicity

Find the derivative (f^\prime(t)=\frac{1}{2\sqrt{t}}+\frac{1}{2\sqrt{1 + t}}). Since (t>0), (\frac{1}{2\sqrt{t}}>0) and (\frac{1}{2\sqrt{1 + t}}>0), so (f^\prime(t)>0) for all (t\in(0,\infty)). A function with a positive derivative on an interval is strictly increasing. A strictly - increasing function can cross the (t) - axis at most once.

Answer:

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