Question

show that the function ( f(t)=sqrt{t}+sqrt{1 + t}-4 ) has exactly one zero in the interval ( (0,infty) ).
solve the equation ( sqrt{t}+sqrt{1 + t}-4 = 0 ) to find the zeros of the given function.
( sqrt{t}+sqrt{1 + t}-4 = 0 )
( sqrt{1 + t}=4-sqrt{t} )
( 1 + t=\frac{225}{64} )

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 \(t
ightarrow\infty\), \(\lim_{t
ightarrow\infty}f(t)=\lim_{t
ightarrow\infty}(\sqrt{t}+\sqrt{1 + t}-4)=\lim_{t
ightarrow\infty}\sqrt{t}(1+\sqrt{1+\frac{1}{t}})-4=\infty>0\).

Step2: Apply the Intermediate Value Theorem

By the Intermediate Value Theorem (since \(f(t)\) is continuous on \((0,\infty)\) and \(f(0)<0\), \(f(t)
ightarrow\infty>0\) as \(t
ightarrow\infty\)), 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 \((0,\infty)\), \(f(0)<0\), \(f(t)
ightarrow\infty\) as \(t
ightarrow\infty\) (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 \((0,\infty)\).