Show that the function f(t)=√t+√√1+t-4 has exactly one zero in Solve the equation √t+√√1+t-4 = 0 to find the zeros of the given √t+√1 +1 -4 = 0 √√1+t=4-√t 1+t=

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem Statement:**

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

**Solution:**

Solve the equation:

\[
\sqrt{t} + \sqrt{1+t} - 4 = 0
\]

to find the zeros of the given function.

Rearranging the equation, we get:

\[
\sqrt{t} + \sqrt{1+t} = 4
\]

Let \( x = \sqrt{t} \), then \( t = x^2 \).

Substitute back into the equation:

\[
x + \sqrt{1+x^2} = 4
\]

Solving for \( x \), we further simplify:

\[
\sqrt{1+x^2} = 4 - x
\]

Squaring both sides, we obtain:

\[
1 + x^2 = (4-x)^2
\]

This equation can be solved for \( x \) to find the zero of the function in the given interval.
Transcribed Image Text:**Problem Statement:** Show that the function \( f(t) = \sqrt{t} + \sqrt{1+t} - 4 \) has exactly one zero in the interval \((0, \infty)\). **Solution:** Solve the equation: \[ \sqrt{t} + \sqrt{1+t} - 4 = 0 \] to find the zeros of the given function. Rearranging the equation, we get: \[ \sqrt{t} + \sqrt{1+t} = 4 \] Let \( x = \sqrt{t} \), then \( t = x^2 \). Substitute back into the equation: \[ x + \sqrt{1+x^2} = 4 \] Solving for \( x \), we further simplify: \[ \sqrt{1+x^2} = 4 - x \] Squaring both sides, we obtain: \[ 1 + x^2 = (4-x)^2 \] This equation can be solved for \( x \) to find the zero of the function in the given interval.
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