数学代写|微积分代写Calculus代写|MATH1111

数学代写|微积分代写Calculus代写|Differentiability implies continuity

One of the phrases defined in definition 1 is “differentiable at $x=k$,” which means the derivative exists at $x=k$. This requires two things: first, that the derivative formula yields the same real result for every infinitesimal $\alpha$; and second, that it does not render $\infty$ or $-\infty$, but renders a real number. This tells the story algebraically but not visually.

Our task for most of the remainder of this section is to determine what differentiability means on the graph of a function.

If $f$ is differentiable at $x=k$, then the function has a tangent line at $x=k$, as in figure 1 . The function must then have a point at $x=k$, meaning that $f(k)$ exists. We have seen the requirement that $f(k)$ exists before-namely, with continuity. Perhaps there is a connection.

Pictures of discontinuities can help us think about differentiability. In figure 2, the function is not defined at the discontinuities and therefore is not differentiable there.What if the value of the function exists at $x=k$, but the function is still discontinuous there? In this case, imagine what the tangent line must look like. The slope of the tangent line is found as the slope of the line through the two points $(k, f(k))$ and $(k+\alpha, f(k+\alpha))$ for an infinitesimal $\alpha$. The second point is located on the curve at a point with the $x$-coordinate essentially the same as $x=k$. Such a line is drawn for various possibilities in figure 3.For an appropriately chosen value of $\alpha$, the result in each graph is a vertical line, and the derivative formula renders either $\infty$ or $-\infty$ for that value of $\alpha$. The function is not differentiable at $x=k$. Although the graphs in figure 3 do not constitute a proof, it is reasonable to conjecture that if a function is not continuous at $x=k$, then it is not differentiable there either.

数学代写|微积分代写Calculus代写|Differentiability, corners, and smoothness

Example 5 Let $f(x)=|x|$. Find $f^{\prime}(0)$.
Solution We use the derivative formula with $k=0$ :
$$f^{\prime}(0)=\frac{f(0+\alpha)-f(0)}{\alpha}=\frac{|\alpha|-|0|}{\alpha}=\frac{|\alpha|}{\alpha}$$ and we are stuck. Because $\alpha$ may be either positive or negative, we do not know how to take its absolute value. When we encountered this same issue with limits, we tried using $\omega$ and $-\omega$ to help us. We can try it here as well.

Using the derivative formula with the positive infinitesimal $\omega$ in place of $\alpha$ gives
$$f^{\prime}(0)=\frac{f(0+\omega)-f(0)}{\omega}=\frac{|\omega|-|0|}{\omega}=\frac{|\omega|}{\omega}=\frac{\omega}{\omega}=1 .$$
We are not stuck with $|\omega|$ because we know that $\omega$ is positive.
Using the derivative formula with the negative infinitesimal $-\omega$ in place of $\alpha$ gives
$$f^{\prime}(0)=\frac{f(0+(-\omega))-f(0)}{-\omega}=\frac{|-\omega|-|0|}{-\omega}=\frac{|-\omega|}{-\omega}=\frac{\omega}{-\omega}=-1 .$$
We are not stuck with $|-\omega|$ because we know that $\omega$ is positive.
Remember that in order for the derivative to exist, we must render the same real result for every infinitesimal $\alpha$, but here we rendered different results when we used the infinitesimals $\omega$ and $-\omega$. The derivative does not exist (DNE); $f^{\prime}(0)$ DNE. The function $f$ is not differentiable at $x=0$.

The results of example 5 are easily explained from the graph of $y=|x|$ (figure 4). When $x$ is positive, $y=|x|$ is the same as $y=x$, which is a line with slope 1 . When $x$ is negative, $y=|x|$ is the same as $y=-x$, which is a line with slope $-1$. This is exactly what we calculated for the slope near zero. Using a positive infinitesimal, the slope is 1 ; using a negative infinitesimal, the slope is $-1$. Where the two lines meet at $(0,0)$ there is a “corner.” At a corner in a graph, the slope on one side of the corner appears to be different from the slope on the other side of the corner, and the function is not differentiable there.

Because a corner in a graph indicates a place where the function is not differentiable, then the graph of a function that is differentiable everywhere must have no corners. We say that such a function is smooth. At the beginning of section $1.6$ we examined the connect-the-dot game with the function $f(x)=x^2$, and saw that the instructions in an algebra course were to connect the points, not by straight line segments, but by making a smooth, connected curve. In example 2, we saw that if $f(x)=x^2$, then $f^{\prime}(x)=2 x$, which is defined everywhere; since $f(x)=x^2$ is differentiable everywhere, then its graph must have no corners. Differentiability is the justification for the instructions you have followed since algebra to draw most functions in a smooth manner. See figure 5.

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数学代写|微积分代写Calculus代写|Differentiability, corners, and smoothness

$$f^{\prime}(0)=\frac{f(0+\alpha)-f(0)}{\alpha}=\frac{|\alpha|-|0|}{\alpha}=\frac{|\alpha|}{\alpha}$$

$$f^{\prime}(0)=\frac{f(0+\omega)-f(0)}{\omega}=\frac{|\omega|-|0|}{\omega}=\frac{|\omega|}{\omega}=\frac{\omega}{\omega}=1 .$$

$$f^{\prime}(0)=\frac{f(0+(-\omega))-f(0)}{-\omega}=\frac{|-\omega|-|0|}{-\omega}=\frac{|-\omega|}{-\omega}=\frac{\omega}{-\omega}=-1 .$$

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