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I have taught high school, community college, and college level math for the past 15 years. I believe that math is easy to learn if you focus on the right concepts, but a lot of education settings make it hard to learn by focusing on the wrong concepts.

Sketch the graph of an example of a function $ f $ that satisfies all of the given conditions.

$ \displaystyle \lim_{x \to 0^-}f(x) = -1 $, $ \displaystyle \lim_{x \to 0^+}f(x) = 2 $, $ f(0) = 1 $

(a) What is wrong with the following equation?$$ \frac {x^2 + x - 6}{x - 2} = x + 3 $$

(b) In view of part (a), explain why the equation $$ \lim_{x \to 2}\frac {x^2 + x - 6}{x - 2} = \lim_{x \to 2}(x + 3) $$is correct.

Use the definition of continuity and the properties of limits to show that the function is continuous at the given number $ a $.

$ f(x) = (x + 2x^3)^4, \hspace{5mm} a = -1 $

Suppose $ f $ is continuous on $ [1, 5] $ and the only solutions of the equation $ f(x) = 6 $ are $ x = 1 $ and $ x = 4 $. If $ f(2) = 8$, explain why $ f(3) > 6 $.

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.

$ x^4 + x - 3 = 0 $, $ (1, 2) $

$ \ln x = x - \sqrt{x} $, $ (2, 3) $