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

show that the function ( f(t)=sqrt{t}+sqrt{1 + t}-4 ) has exactly one zero in the interval ( (0,infty) ).\nsolve the equation ( sqrt{t}+sqrt{1 + t}-4 = 0 ) to find the zeros of the given function.\n\begin{align*}sqrt{t}+sqrt{1 + t}-4&=0\\sqrt{1 + t}&=4-sqrt{t}\\1 + t&=16-8sqrt{t}+t\\8sqrt{t}&=15\\t&=\frac{225}{64}quad(\text{simplify your answer.})end{align*}\nthe function ( f(t)=sqrt{t}+sqrt{1 + t}-4 ) has a zero at ( t=\frac{225}{64} ), which is in the interval ( (0,infty) ).\nrolles 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^{prime}(c)=0 ).\nfind the derivative of ( f(t)=sqrt{t}+sqrt{1 + t}-4 ).\n( f^{prime}(t)=square )

show that the function ( f(t)=sqrt{t}+sqrt{1 + t}-4 ) has exactly one zero in the interval ( (0,infty) ).\nsolve the equation ( sqrt{t}+sqrt{1 + t}-4 = 0 ) to find the zeros of the given function.\n\begin{align*}sqrt{t}+sqrt{1 + t}-4&=0\\sqrt{1 + t}&=4-sqrt{t}\\1 + t&=16-8sqrt{t}+t\\8sqrt{t}&=15\\t&=\frac{225}{64}quad(\text{simplify your answer.})end{align*}\nthe function ( f(t)=sqrt{t}+sqrt{1 + t}-4 ) has a zero at ( t=\frac{225}{64} ), which is in the interval ( (0,infty) ).\nrolles 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^{prime}(c)=0 ).\nfind the derivative of ( f(t)=sqrt{t}+sqrt{1 + t}-4 ).\n( f^{prime}(t)=square )

Answer

Explanation:

Step1: Differentiate (\sqrt{t})

Using the power rule ((x^n)^\prime = nx^{n - 1}), for (y=\sqrt{t}=t^{\frac{1}{2}}), (y^\prime=\frac{1}{2}t^{\frac{1}{2}-1}=\frac{1}{2\sqrt{t}})

Step2: Differentiate (\sqrt{1 + t})

Let (u = 1 + t), then (y=\sqrt{u}=u^{\frac{1}{2}}). By the chain rule (\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}). (\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}) and (\frac{du}{dt}=1). So (\frac{d}{dt}(\sqrt{1 + t})=\frac{1}{2\sqrt{1 + t}})

Step3: Differentiate the constant (-4)

The derivative of a constant (C) is (0). So (\frac{d}{dt}(-4)=0)

Step4: Find (f^\prime(t))

Using the sum rule ((u + v+w)^\prime=u^\prime + v^\prime+w^\prime), where (u = \sqrt{t}), (v=\sqrt{1 + t}), (w=-4). Then (f^\prime(t)=\frac{1}{2\sqrt{t}}+\frac{1}{2\sqrt{1 + t}}+0=\frac{1}{2\sqrt{t}}+\frac{1}{2\sqrt{1 + t}})

Answer:

(f^\prime(t)=\frac{1}{2\sqrt{t}}+\frac{1}{2\sqrt{1 + t}})