How to Show f(x) is Close to L Using Epsilon-Delta Proof

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To demonstrate that f(x) is close to L using an epsilon-delta proof, start by rewriting the inequality |f(x)-2| < 0.05 as 1.95 < f(x) < 2.05. This leads to finding the x-values for which f(x) equals 1.95 and 2.05, which can be done using a graphing utility. The graph will help visualize the function's behavior around the point of interest, aiding in estimating delta. Understanding the relationship between epsilon and delta is crucial for grasping the concept of closeness in calculus.
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Generate the graph of f(x)=x^3-4x+5 with a graphing utility. And use the graph to find delta such that |f(x)-2|<.05 if 0<|x-1|<delta [Hint show that the inequality |f(x)-2| < .05 can be rewritten 1.95 < x^3-4x+5 < 2.05, then estimate the values of x for which f(x)=1.95 and f(x)=2.05]. Now I can do the graphing parts and I can correctly estimate the values of x, but I cannot do the "show that the inequality |f(x)-2|<.05 can be rewritten etc." Can anyone show me how to do it as I have started studying a University maths book on my own. Thanks.
 
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|f(x) - L| < e means that f(x) is in (L - e, L + e) which means that L - e < f(x) < L + e. Also, I'm not sure why the text asks you to use a graphing utility since it's maybe instructive once or twice to see an epsilon-delta proof done with an explicit (numerical) epsilon to get a feel what is meant by "closeness" of f(x) to L or whatever.
 
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