The function f(x) = −2x² + 4x³ 3 has two critical values x = 0, and x = on the interval 2 [-7,7] A) First critical value, x = 0, is A relative minimum, at value f(0): = 0 Neither a relative maximum nor a relative minimum, at value f(0) = 0 A relative maximum, at value f(0) = 0 Not enough information to determine 3 B) Second critical value, x = is 2 Neither a relative maximum nor a relative 3 minimum, at value f 2 = 27 8 A relative minimum, at value f 32 (33) A relative maximum, at value ƒ ( 3 2 Not enough information to determine = 27 8 = 27 8 C) Find the Absolute Maximum: 27 Absolute Maximum is at x = 8 2 N/W 3 27 Absolute Maximum is at x = 2 8 Absolute Maximum is at more than one x values Not enough information to determine D) Find the Absolute Minimum: Absolute Minimum is -6174 at x = -7. Absolute Minimum is at more than one x values Absolute Minimum is -7 at x=-6174. Not enough information to determine

The function f(x) = −2x² + 4x³ 3 has two critical values x = 0, and x = on the interval 2 [-7,7] A) First critical value, x = 0, is A relative minimum, at value f(0): = 0 Neither a relative maximum nor a relative minimum, at value f(0) = 0 A relative maximum, at value f(0) = 0 Not enough information to determine 3 B) Second critical value, x = is 2 Neither a relative maximum nor a relative 3 minimum, at value f 2 = 27 8 A relative minimum, at value f 32 (33) A relative maximum, at value ƒ ( 3 2 Not enough information to determine = 27 8 = 27 8 C) Find the Absolute Maximum: 27 Absolute Maximum is at x = 8 2 N/W 3 27 Absolute Maximum is at x = 2 8 Absolute Maximum is at more than one x values Not enough information to determine D) Find the Absolute Minimum: Absolute Minimum is -6174 at x = -7. Absolute Minimum is at more than one x values Absolute Minimum is -7 at x=-6174. Not enough information to determine

College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter4: Polynomial And Rational Functions

Section4.2: Polynomial Functions

Problem 96E: What is the purpose of the Intermediate Value Theorem?

See similar textbooks

Similar questions

What is the purpose of the Intermediate Value Theorem?
How are the absolute maximum and minimum similar to and different from the local extrema?
find the critical points for the function f(x)=2x³ – 3x2 – 72x-11 оз, -4 O None о -3, -3,4 O -1,-5
The function f(x) is increasing on the open interval ( 4x³ +7.56x² + 366.2004x + 7.99 ). = It is decreasing on the open interval (-∞, The function has a local maximum at " ) and the open interval ( , ∞ ).
1) Consider f(x) = 3x4 -8x³ +6 a) Find all critical points of the function (make sure to consider if the derivative is defined everywhere or not!). Label them in increasing order, 21, 22, 23, etc (for how many you find). b) Use the first derivative test to determine if x₁ is a local max, min, or neither. Find the y value associated with this a value and write it as a point (x, y). 1 c) Use the second derivative test to determine if x2 is a local max, min, or neither. Find the y value associated with this a value and write it as a point (x, y).
Find the x-coordinates of all critical points of the given function. Determine whether each critical point is a relative maximum, minimum, or neither by first applying the second derivative test, and, if the test fails, by some other method. f(x) = 5x6 - 14x4 f has ---Select--- v at the critical point x = (smaller x-value). The second derivative test --Select--- v for this critical point. f has ---Select--- at the critical point x = (larger x-value). The second derivative test --Select-- v for this critical point. f has --Select--- v at the critical point x = (largest x-value). The second derivative test -Select--- v for this critical point.
The stock price for a certain corporation can be approximated for the years 2012 through 2017 by f(x)% 3D6.3x- 168x+ 1199, where x = 12 corresponds to the year 2012. During what year did a local minimum stock price occur? A local extremum occurs at a critical number of f. Because the derivative exists for every x, the only critical number(s) occur where the derivative is zero. Find the derivative of f(x). f'(x) =
8. (a) State the Intermediate Value Theorem and use it to show that y = r³ + x +1 has exactly one real zero. (b) State the Mean Value Theorem and use it to show that f(x) critical point. Find that critical point. = xe-* has at least one
Find the critical points of y=(16-2x)(30-2x)x in an interval [0,8]