Question

1. Let $f: [0,1] \to \mathbb{R}$ be a continuous function and define $g: [0,1] \to \mathbb{R}$ by $g(x) = \max\{f(y): 1 \le y \le x\}$, $x \in [0,1]$. Show that $g$ is continuous on $[0, 1]$.

          1. Let $f: [0,1] \to \mathbb{R}$ be a continuous function and define $g: [0,1] \to \mathbb{R}$ by

$g(x) = \max\{f(y): 1 \le y \le x\}$, $x \in [0,1]$.

Show that $g$ is continuous on $[0, 1]$.

1. Let f: [0,1] →ℝ be a continuous function and define g: [0,1] →ℝ by

g(x) = max{f(y): 1 ≤ y ≤ x}, x ∈ [0,1].

Show that g is continuous on [0, 1].

Added by Carla S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

07/06/2023

Step 1

Step 1: To show that $g$ is continuous on $[0,1]$, we need to show that for any $x_0 \in [0,1]$ and any $\epsilon > 0$, there exists a $\delta > 0$ such that $|g(x) - g(x_0)| < \epsilon$ whenever $|x - x_0| < \delta$. Show more…

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1. Let $f: [0,1] \rightarrow \mathbb{R}$ be a continuous function and define $g: [0,1] \rightarrow \mathbb{R}$ by $g(x) = max\{f(y): 1 \leq y \leq x\}$, $x \in [0,1]$. Show that $g$ is continuous on $[0, 1]$.