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

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

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

Answer

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