- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
4. Applications of Derivatives
Differentials
Problem 2.82.2
Textbook Question
b. Estimate a solution to the equation in the given interval using a root finder.
x^5+7x+5=0; (−1,0)

1
First, understand that the problem requires finding a root of the equation \(x^5 + 7x + 5 = 0\) within the interval \((-1, 0)\). A root finder method, such as the bisection method, can be used for this purpose.
Begin by evaluating the function at the endpoints of the interval. Calculate \(f(-1)\) and \(f(0)\) to check if there is a sign change, which indicates the presence of a root in the interval.
If there is a sign change, apply the bisection method: calculate the midpoint of the interval, \(c = \frac{-1 + 0}{2} = -0.5\), and evaluate \(f(c)\).
Determine which subinterval, \((-1, -0.5)\) or \((-0.5, 0)\), contains a sign change by checking the sign of \(f(c)\) and comparing it with the signs of \(f(-1)\) and \(f(0)\).
Repeat the bisection process on the subinterval with the sign change, continually halving the interval and evaluating the function at the new midpoints, until the interval is sufficiently small to estimate the root accurately.
Recommended similar problem, with video answer:

This video solution was recommended by our tutors as helpful for the problem above
Video duration:
3mPlay a video:
Was this helpful?