Concept explainers
In Exercises 1–8, find all the relative and absolute extrema of the given function on the given domain (if supplied) or on the largest possible domain (if no domain is supplied).
To calculate: The exact location of the relative and absolute extrema of the function
Answer to Problem 1RE
Solution:
The exact location of Absolute Minimum are
Explanation of Solution
Given Information:
The provided equation is:
Formula used:
First Derivative Test is, assume that for a critical point
If
If
If
Stationary Points are the points in the interior of the domain where the derivative is zero.
Singular Points are the points in the interior of the domain where the derivative is not defined.
Closed intervals contain end points which are the end points of the functions and the function does not have end points if the interval is open as open intervals do not have any end points.
Calculation:
Consider the provided equation:
Differentiate both sides of the equation with respect to
Now, locate stationary points.
Recall that stationary Points are the points in the interior of the domain where the derivative is zero.
Adding
Multiplying both sides of the equation by
Taking square root of both sides of the equation to get:
The domain of the function is
Now locate singular points.
Since
Next locate end points.
Since
The values of
5 | 5 |
The graph increases from
Use the above points to plot a graph
From figure we can see that
Classification | ||
Absolute Minimum | ||
Absolute Minimum | ||
Relative Maximum |
Therefore, the exact location of Absolute Minimum are
Want to see more full solutions like this?
Chapter 5 Solutions
Applied Calculus
- Find the absolute maximum and absolute minimum of f(x)=e−2xf(x)=e−2x over the interval [−1,3][−1,3].arrow_forwardIn Exercises 27–32, use a graphingutility to graph the function on the closed interval [a, b].Determine whether Rolle’s Theorem can be applied to f on theinterval and, if so, find all values of c in the open interval (a, b)such that f '(c= ' 0.) f(x)=|x|-1,[-1,1]arrow_forwardIn Exercises 45-4-6, we demonstrate that, in general, you cannot obtain anantiderivative of a product of functions by taking a product of antiderivatives of each. Show that G(x) = x 2ex is not an antiderivative of f(x) = 2xex butH(x) = 2xex - 2ex is.arrow_forward
- In Exercises 39 and 40, what value should be as-signed to k to make f a continuous function?arrow_forwardFind the absolute maximum and absolute minimum values of f on the given interval. f(x) = x/x2+1 , [0, 3]arrow_forwardOn what open interval is f ' an increasing function? For which value of x is f '(x) minimum?arrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning