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.
\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}
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^{prime}(c)=0 ).
find the derivative of ( f(t)=sqrt{t}+sqrt{1 + t}-4 ).
( f^{prime}(t)=square )

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:

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