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Suppose fx=mx+bis a linear function with m≠0, and let c be any real member.

Part (a): Show that for all ∈>0, if 0<x-c<∈m, then fx-fc<∈.

Part (b): What does the implication in part (a) have to do with limits?

Part (c): Illustrate the implication in part (a) with a labeled graph. Explain in terms of slopes why it makes sense that given ∈>0, the correspondingδ>0isδ=∈m.

Short Answer

Expert verified

Part (a): To prove fx-fc<∈, put x=cand multiply mon every side and solve the inequation.

Part (b): Using limits, we get limx→cx=c.

Part (c): The required graph is given below,

Step by step solution

01

Part (a) Step 1. Given information.

Consider the given question,

fx=mx+b∈>0

If 0<x-c<∈mthenfx-fc<∈.

02

Part (a) Step 2. Substitute c in x in the given function.

Substitute x=cin the given function,

fc=mc+c

For all ∈>0,

0<x-c<∈m

Multiply mon every side,

role="math" localid="1648275234533" 0·m<x-c·m<∈m·m0<mx-mc<∈0<mx+c-mc+c<∈0<fx-fc<∈

Therefore, fx-fc<∈.

Hence, proved.

03

Part (b) Step 1. To defined the given implication with limits.

From the definition of limit,

If 0<x-c<∈m,

Then, it can be written,

limx→cx=c

04

Part (c) Step 1. Sketch the graph.

On sketching the graph, we get,

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