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# Let$$f(x) = \left\{ \begin{array}{ll} |x|^x & \mbox{if} x \not= 0\\ 1 & \mbox{if} x = 0\\ \end{array} \right.$$(a) Show that $f$ is continuous at $0$.(b) Investigate graphically whether $f$ is differentiable at $0$ by zooming in several times toward the point $(0, 1)$ on the graph of $f$.(c) Show that $f$ is not differentiable at $0$. How can you reconcile this fact with the appearance of the graphs in part (b)?

## a)Find the limit as you approach 0 from two sides, if this limit equals $f(0)$then it is continuous at $0 .$b) It looks like it should be differentiable at $x=0$c) Now, as $x \rightarrow 0, f^{\prime}(x) \rightarrow-\infty,$ so there is a vertical tangent at $(0,1)$ and the function is not differentiable at this point.

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for this problem, we're wanting to show um that F is going to be continuous And it's gonna be continuous at zero. So we can look at this by showing the absolute value of X raised to the power of X. We know that the continuous function because for all um X. That's going to equal a value. But we say that that can't be the case if X equals zero. So it's important for us to determine what that value is and we see that X equals zero. The function equals one. So that's good because the second portion of the graph is just the line one. Um that's why equals one only at that point. So because of that, we see the graph is going to be continuous because the limit As this graph, the purple graph, the limit as the purple graph approaches zero is 1. And since one is the value here, we see that the condition is satisfied, meaning that is going to be a continuous function. Um So based on that, we could use local tiles rule to show that this function here as X approaches zero equals one. We could also do it conceptually by thinking about it or graphically as well.

California Baptist University

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