Let g be a continuous function on the closed interval [0, 1]. Let g(0) = 1 and g(1) = 0. Which...

Question:

Let {eq}g {/eq} be a continuous function on the closed interval {eq}[0, 1] {/eq}. Let {eq}g(0) = 1 {/eq} and {eq}g(1) = 0 {/eq}.

Which of the following is NOT necessarily true?

(a) There exists a number {eq}h {/eq} in {eq}[0, 1] {/eq} such that {eq}g(h) \geq g(x) {/eq} for all {eq}x {/eq} in {eq}[0, 1] {/eq}.

(b) For all {eq}a {/eq} and {eq}b {/eq} in {eq}[0, 1] {/eq}, if {eq}a = b {/eq}, then {eq}g(a) = g(b) {/eq}.

(c) There exists a number {eq}h {/eq} in {eq}[0, 1] {/eq} such that {eq}g(h) = \frac{1}{2} {/eq}.

(d) There exists a number {eq}h {/eq} in {eq}[0, 1] {/eq} such that {eq}g(h) = \frac{3}{2} {/eq}.

(e) For all {eq}h {/eq} in the open interval {eq}(0, 1) , lim_{x \rightarrow h} g(x) = g(h) {/eq}.

Theorems for Continuous Functions:

The intermediate value theorem says that for any continuous function {eq}f(x) {/eq}, if {eq}d {/eq} is any value in between {eq}f(a) {/eq} and {eq}f(b) {/eq}, then there exists some {eq}c {/eq} such that {eq}f(c)=d {/eq}.

The extreme value theorem says that any function {eq}f(x) {/eq} on the closed interval {eq}[a, b] {/eq} has a maximum and a minimum value on that interval.

Answer and Explanation: 1

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Statement (a) is true by the extreme value theorem. It is just another way of saying that {eq}g(x) {/eq} has a maximum value on the interval {eq}[0,...

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Continuous Functions Theorems

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Chapter 5 / Lesson 10
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Continuous functions are functions that have their conditions satisfied between multiple points, appearing as an uninterrupted line when graphed. See examples of how this is represented in the intermediate value theorem and the extreme value theorem.


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