Worksheet12: Dickson 059
Current Work (editable)
is denoted by f^{(k)}(x). Thus (6) f'(x) = na_{0} x^{n-1} + (n - 1)a_{1} x^{n-2} + ... + 2a_{n-2} x + a_{n-1}, (7) f''(x) = n(n - 1)a_{0} x^{n-2} + (n - 1)(n - 2)a_{1} x^{n-3} + ... + 2a_{n-2}, etc. Hence we have (8) f(x + h) = f(x) + f'(x)h + f''(x) h^{2}/1·2 + f'''(x) h^{3}/1·2·3 + ... + f^{(r)}(x) h^{r}/r! + ... + f^{(n)}(x) h^{n}/n!, where r! is the symbol, read r factorial, for the product 1·2·3 ... (r - 1)r. Here r is a positive integer, but we include the case r = 0 by the definition, 0! = 1. This formula (8) is known as Taylor's theorem for the present case of a polynomial f(x) of degree n. We call f'(x) the (first) derivative of f(x), and f''(x) the second derivative of f(x), etc. Concerning the fact that f''(x) is equal to the first derivative of f'(x) and that, in general, the kth derivative f^{(k)}(x) of f(x) is equal to the first derivative of f^{(k-1)}(x), see Exs. 6-9 of the next set. In view of (8), the limit of (4) as h approaches zero is f'(x). Hence f'(x) is the slope of the tangent to the graph of y = f(x) at the point (x, y). In (5) and (6), let every a be zero except a_{0}. Thus the derivative of a_{0} x^{n} is na_{0} x^{n-1}, and hence is obtained by multiplying the given term by its exponent n and then diminishing its exponent by unity. For example, the derivative of 2x^{3} is 6x^{2}. Moreover, the derivative of f(x) is equal to the sum of the derivatives of its separate terms. Thus the derivative of x^{3} + 4x^{2} - 11 is 3x^{2} + 8x, as found also in §55. EXERCISES 1. Show that the slope of the tangent to y = 8x^{3} - 22x^{2} + 13x - 2 at (x, y) is 24x^{2} - 44x + 13, and that the bend points are (0.37, 0.203), (1.46, -5.03), approximately. Draw the graph. 2. Prove that the bend points of y = x^{3} - 2x - 5 are (.82, -6.09), (-.82, -3.91), approximately. Draw the graph and locate the real roots. 3. Find the bend points of y = x^{3} + 6x^{2} + 8x + 8. Locate the real roots. 4. Locate the real roots of f(x) = x^{4} + x^{3} - x - 2 = 0. Hints: The abscissas of the bend points are the roots of f'(x) = 4x^{3} + 3x^{2} - 1 = 0. The bend points of y = f'(x) are (0, -1) and (-1/2, -3/4), so that f'(x)= 0 has a single real root (it is just less than 1/2). The single bend point of y = f(x) is (1/2, -37/16), approximately.
Answer Key (non-editable)
is denoted by $f^{(k)}(x)$. Thus \begin{align*} \Tag{(6)} f'(x) &= na_{0} x^{n-1} + (n - 1)a_{1} x^{n-2} + \dots + 2a_{n-2} x + a_{n-1}, \\ \Tag{(7)} f''(x) &= n(n - 1)a_{0} x^{n-2} + (n - 1)(n - 2)a_{1} x^{n-3} + \dots + 2a_{n-2}, \end{align*} etc. Hence we have \begin{multline*} \Tag{(8)} f(x + h) = f(x) + f'(x)h + f''(x) \frac{h^{2}}{1·2} + f'''(x) \frac{h^{3}}{1·2·3} \\ + \dots + f^{(r)}(x) \frac{h^{r}}{r!} + \dots + f^{(n)}(x) \frac{h^{n}}{n!}, \end{multline*} where $r!$ is the symbol, read \emph{$r$ factorial}, for the product $1·2·3 \dots (r - 1)r$. Here $r$ is a positive integer, but we include the case $r = 0$ by the definition, $0! = 1$. This formula \Eq{(8)} is known as \emph{Taylor's theorem} for the present case of %[** Attn italic (), inconsistent ital.] a polynomial $f(x)$ of degree $n$. We call $f'(x)$ the \emph{(first) derivative of $f(x)$}, and $f''(x)$ the \emph{second derivative} of $f(x)$, etc. Concerning the fact that $f''(x)$ is equal to the first derivative of $f'(x)$ and that, in general, the $k$th derivative $f^{(k)}(x)$ of $f(x)$ is equal to the first derivative of $f^{(k-1)}(x)$, see Exs.~6--9 of the next set. In view of \Eq{(8)}, the limit of \Eq{(4)} as $h$ approaches zero is $f'(x)$. Hence \begin{Thm}% $f'(x)$ is the slope of the tangent to the graph of $y = f(x)$ at the point $(x, y)$. \end{Thm} In \Eq{(5)} and \Eq{(6)}, let every $a$ be zero except $a_{0}$. Thus the derivative of $a_{0} x^{n}$ is $na_{0} x^{n-1}$, and hence is obtained by multiplying the given term by its exponent $n$ and then diminishing its exponent by unity. For example, the derivative of $2x^{3}$ is $6x^{2}$. Moreover, the derivative of $f(x)$ is equal to the sum of the derivatives of its separate terms. Thus the derivative of $x^{3} + 4x^{2} - 11$ is $3x^{2} + 8x$, as found also in §55. \begin{Exercises} \begin{itemize} \item[1.] Show that the slope of the tangent to $y = 8x^{3} - 22x^{2} + 13x - 2$ at $(x, y)$ is $24x^{2} - 44x + 13$, and that the bend points are $(0.37, 0.203)$, $(1.46, -5.03)$, approximately. Draw the graph. \item[2.] Prove that the bend points of $y = x^{3} - 2x - 5$ are $(.82, -6.09)$, $(-.82, -3.91)$, approximately. Draw the graph and locate the real roots. \item[3.] Find the bend points of $y = x^{3} + 6x^{2} + 8x + 8$. Locate the real roots. \item[4.] Locate the real roots of $f(x) = x^{4} + x^{3} - x - 2 = 0$. Hints: The abscissas of the bend points are the roots of $f'(x) = 4x^{3} + 3x^{2} - 1 = 0$. The bend points of $y = f'(x)$ are $(0, -1)$ and $(-\frac{1}{2}, -\frac{3}{4})$, so that $f'(x)= 0$ has a single real root (it is just less than $\frac{1}{2}$). The single bend point of $y = f(x)$ is $(\frac{1}{2}, -\frac{37}{16})$, approximately. \end{Exercises}%[** Continues]
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