Question

show that the function $f(t)=\sqrt{t}+\sqrt{1 + t}-4$ has exactly one zero in the interval $(0,\infty)$.the function $f(t)=\sqrt{t}+\sqrt{1 + t}-4$ has a zero at $t = \frac{225}{64}$, which is in the interval $(0,\infty)$.rolles theorem states that for a function $f(x)$ that is continuous at every point over the closed interval $a,b$ and differentiable at every point of its interior $(a,b)$, if $f(a)=f(b)$, then there is at least one number $c$ in $(a,b)$ at which $f(c)=0$.find the derivative of $f(t)=\sqrt{t}+\sqrt{1 + t}-4$.$f(t)=\frac{1}{2\sqrt{t}}+\frac{1}{2\sqrt{1 + t}}$can the derivative of $f(t)$ be zero in the interval $(0,\infty)$?yesno

Explanation:

Step1: Analyze the derivative

For \(f(t)=\sqrt{t}+\sqrt{1 + t}-4\), its derivative \(f^{\prime}(t)=\frac{1}{2\sqrt{t}}+\frac{1}{2\sqrt{1 + t}}\).
Since \(t>0\), \(\sqrt{t}>0\) and \(\sqrt{1 + t}>0\). Then \(\frac{1}{2\sqrt{t}}>0\) and \(\frac{1}{2\sqrt{1 + t}}>0\).

Step2: Determine if the derivative can be zero

So \(f^{\prime}(t)=\frac{1}{2\sqrt{t}}+\frac{1}{2\sqrt{1 + t}}>0\) for all \(t\in(0,\infty)\).

Answer:

B. No