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