## 数学代写|微积分代写Calculus代写|Vertical tangents

There is yet one more feature on a graph that can indicate nondifferentiability.
Example 6 Let $f(x)=x^{1 / 3}$. Find $f^{\prime}(0)$.
Solution We use the derivative formula with $k=0$ :
\begin{aligned} f^{\prime}(0) &=\frac{f(0+\alpha)-f(0)}{\alpha}=\frac{\alpha^{1 / 3}-0^{1 / 3}}{\alpha} \ &=\frac{\alpha^{1 / 3}}{\alpha}=\frac{1}{\alpha^{2 / 3}}=A^{2 / 3} \ & \doteq \infty . \end{aligned}
We are not stuck at $A^{2 / 3}$ because of the even numerator in the exponent; $A^{2 / 3}=\left(A^2\right)^{1 / 3}$, and $A^2$ is positive. We can render $\infty$. Because we rendered $\infty$ rather than a real number, $f^{\prime}(0)$ does not exist.

Since $f^{\prime}(0)=\infty$, it seems the slope of the tangent line at $x=0$ is positive infinite. This is true, as verified by the graph; see figure 6 .
The procedure for determining the equation of the tangent line to the curve $y=x^{1 / 3}$ at $x=0$ is not the same as before. Because the equation of a vertical line is of the form $x=k$, we cannot use the pointslope equation of the line. Instead, we simply use the $x$-coordinate at which we are finding the derivative; the equation of the tangent line is $x=0$.

If the derivative renders an infinite value, the tangent line is vertical. Figure 7 shows four of the possibilities.The two graphs on the right contain a cusp, where there is a corner with a vertical tangent line. The cusp is a point on the graph; the function is defined there. Because of the manner in which calculators and computer software calculate and display graphs of functions, it is sometimes difficult to distinguish cusps from vertical asymptotes using only a graph, but calculus can help you determine the difference!

## 数学代写|微积分代写Calculus代写|Proofs of theorems

The local linearity formula is part of the local linearity theorem.
Theorem 2 LOCAL LINEARITY THEOREM Let $f$ be differentiable at $x=k$ and let $\alpha$ be an infinitesimal.

(1) If not both $f(k)=0$ and $f^{\prime}(k)=0$, then $f(k+\alpha) \approx f(k)+\alpha f^{\prime}(k)$.
(2) If $f(k)=0=f^{\prime}(k)$, then $f(k+\alpha)$ is zero or is on a lower level than $\alpha$.
Proof. (1) Case 1: $f^{\prime}(k) \neq 0$.
Since $f$ is differentiable at $k, \frac{f(k+\alpha)-f(k)}{\alpha} \doteq f^{\prime}(k)$. Since $f^{\prime}(k) \neq 0$ is real, $\frac{f(k+\alpha)-f(k)}{\alpha} \approx f^{\prime}(k)$, and in fact $\frac{f(k+\alpha)-f(k)}{\alpha}=f^{\prime}(k)+\beta$, where $\beta$ is zero or infinitesimal. Then, $f(k+\alpha)-f(k)=\alpha f^{\prime}(k)+\alpha \beta$, and $f(k+\alpha)=f(k)+\alpha f^{\prime}(k)+\alpha \beta$. Because $\alpha \beta$ is either zero or is on a lower level than $\alpha f^{\prime}(k)$, this term can be discarded and $f(k+\alpha) \approx$ $f(k)+\alpha f^{\prime}(k)$
(1) Case 2: $f(k) \neq 0, f^{\prime}(k)=0$.
Then, since $\frac{f(k+\alpha)-f(k)}{\alpha} \doteq f^{\prime}(k)=0, \frac{f(k+a)-f(k)}{\alpha}=\beta$, where $\beta$ is zero or infinitesimal. Then, $f(k+\alpha)=f(k)+\alpha \beta \approx f(k)=f(k)+\alpha f^{\prime}(k)$.
(2) Now suppose $f^{\prime}(k)=f(k)=0$. Then, $\frac{f(k+a)-f(k)}{\alpha}=\frac{f(k+\alpha)}{\alpha} \doteq 0$, so $\frac{f(k+\alpha)}{\alpha}$ is zero or infinitesimal. Thus, either $f(k+\alpha)=0$ or $f(k+\alpha)$ is on a lower level than $\alpha$.

It remains to furnish a complete proof of theorem 1. Let $f$ be a function that is differentiable at $x=k$. We wish to prove that $f$ is continuous at $x=k$.

Proof. (theorem 1) Case (1): $f(k) \neq 0$. By the local linearity theorem part $(1), f(k+\alpha) \approx f(k)+\alpha f^{\prime}(k) \approx f(k)$. By definition, $f$ is continuous at $x=k$.

Case (2): $f(k)=0, f^{\prime}(k) \neq 0$. By the local linearity theorem part (1), $f(k+\alpha) \approx f(k)+\alpha f^{\prime}(k)=\alpha f^{\prime}(k) \doteq 0=f(k)$. By definition, $f$ is continuous at $x=k$.

Case (3): $f(k)=0, f^{\prime}(k)=0$. By the local linearity theorem part (2), $f(k+\alpha) \doteq 0=f(k)$. By definition, $f$ is continuous at $x=k$.

Formal proofs that involve the local linearity formula $f(k+\alpha) \approx$ $f(k)+\alpha f^{\prime}(k)$ nearly always require the use of both parts of the local linearity theorem. In this text, results are usually derived using the local linearity formula only and the rest of the details are left unstated.

# 微积分代考

## 数学代写|微积分代写Calculus代写|Vertical tangents

$$f^{\prime}(0)=\frac{f(0+\alpha)-f(0)}{\alpha}=\frac{\alpha^{1 / 3}-0^{1 / 3}}{\alpha} \quad=\frac{\alpha^{1 / 3}}{\alpha}=\frac{1}{\alpha^{2 / 3}}=A^{2 / 3} \doteq \infty$$

## 数学代写|微积分代写Calculus代写|Proofs of theorems

(1) 如果不是两者 $f(k)=0$ 和 $f^{\prime}(k)=0$ ，然后 $f(k+\alpha) \approx f(k)+\alpha f^{\prime}(k)$.
(2) 如果 $f(k)=0=f^{\prime}(k)$ ，然后 $f(k+\alpha)$ 为雴或低于 $\alpha$.

$(-)$ 案例二: $f(k) \neq 0, f^{\prime}(k)=0$.

(2) 现在假设 $f^{\prime}(k)=f(k)=0$. 然后， $\frac{f(k+a)-f(k)}{\alpha}=\frac{f(k+\alpha)}{\alpha} \doteq 0$ ，所以 $\frac{f(k+\alpha)}{\alpha}$ 为零或无穷小。因此， 无论是 $f(k+\alpha)=0$ 或者 $f(k+\alpha)$ 低于 $\alpha$.

$f(k+\alpha) \approx f(k)+\alpha f^{\prime}(k)=\alpha f^{\prime}(k) \doteq 0=f(k)$. 根据定义, $f$ 是连续的 $x=k$.

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