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$$ \begin{array}{lcl} (a^2 + b^2)^3 & \ne & a^6 + b^6\\

(a^2 + b^2)^3 & = & (a^2 + b^2) (a^2 + b^2) (a^2 + b^2)\\

& = & (a^4 + 2a^2b^2 + b^4) (a^2 + b^2)\\

& = & (a^6 + 2a^4b^2 + a^2b^4 + a^4b^2 + 2a^2b^4 + b^6) \end{array} $$

63. Suppose that L is the tangent line at $$x=x_0$$ to the graph of the cubic equation $$y = ax^3 + bx$$. Find the x-coordinate of the point where L intersects the graph a second time.

First, we find the derivative of y:

$$y' = 3ax^2 + b\,$$

This gives us the slope of L at $$x_0$$. Next we designate the point $$(x_0,y_0)$$ as the point where L is tangent to y(Note: $$x_0$$ is treated as a constant from here on out).

Now, using the point-slope form of a line, we define L:

$$L = y - y_0 = (3ax_0^2 + b)(x - x_0)$$

We can write $$y_0$$ in terms of $$x_0$$ using the original equation:

$$y_0 = ax_0^3 + bx_0$$

Then,

$$ \begin{array}{lcl}

L = y - (ax_0^3 + bx_0) & = & (3ax_0^2 + b)(x - x_0) \\

y - ax_0^3 - bx_0 & = & 3ax_0^2x - 3ax_0^3 + bx - bx_0\\

y & = & 3ax_0^2x - 2ax_0^3 + bx

\end{array} $$

Now that we have the above formula for the tangent line L, we set it equal to the original cubic equation and find all the solutions:

$$ \begin{array}{lcl}

ax^3 + bx & = & 3ax_0^2x - 2ax_0^3 + bx\\

ax^3 & = & 3ax_0^2x - 2ax_0^3\\

x^3 & = & 3x_0^2x - 2x_0^3\\

x^3 -3x_0^2x + 2x_0^3 & = & 0

\end{array} $$

To factor the above we will use synthetic division. We already know that $$x-x_0$$ is a factor, because $$x_0$$ is where L is tangent to the above.

$$x - x_0\overline{\vert x^3 + 0x^2 - 3x_0^2x + 2x_0^3}$$

$$\begin{matrix} x_0 & | & 1 & 0 & -3x_0^2 & 2x_0^3\\ & | & & & & \\ & | & 1 & & & \\ \end{matrix}$$

$$\begin{matrix} x_0 & | & 1 & 0 & -3x_0^2 & 2x_0^3\\ & | & & x_0 & & \\ & | & 1 & & & \\ \end{matrix}$$

$$\begin{matrix} x_0 & | & 1 & 0 & -3x_0^2 & 2x_0^3\\ & | & & x_0 & & \\ & | & 1 & x_0 & &\\ \end{matrix}$$

$$\begin{matrix} x_0 & | & 1 & 0 & -3x_0^2 & 2x_0^3\\ & | & & x_0 & x_0^2 & \\ & | & 1 & x_0 & &\\ \end{matrix}$$

$$\begin{matrix} x_0 & | & 1 & 0 & -3x_0^2 & 2x_0^3\\ & | & & x_0 & x_0^2 & \\ & | & 1 & x_0 & -2x_0^2 & \\ \end{matrix}$$

$$\begin{matrix} x_0 & | & 1 & 0 & -3x_0^2 & 2x_0^3\\ & | & & x_0 & x_0^2 & -2x_0^3 \\ & | & 1 & x_0 & -2x_0^2 & \\ \end{matrix}$$

$$\begin{matrix} x_0 & | & 1 & 0 & -3x_0^2 & 2x_0^3\\ & | & & x_0 & x_0^2 & -2x_0^3 \\ & | & 1 & x_0 & -2x_0^2 & 0\\ \end{matrix}$$

Since the remainder is 0, this confirms that $$x-x_0$$ is a factor.

$$ \begin{array}{lcl} (x^2 + x_0x - 2x_0^2)(x - x_0) & = & 0\\ (x - x_0)(x + 2x_0)(x - x_0) & = & 0\\ (x - x_0)^2(x + 2x_0) & = & 0\\ \end{array} $$

Thus, L crosses $$y = ax^3 + bx\,$$ at $$ x = -2x_0\, $$



Given:

$$ \begin{align} f'(x) & = \lim_{h \to 0}{f(x+h)-f(x)\over(h)} \\ f(x) & = x^n \end{align} $$

Prove:

$$ f'(x) = nx^{n-1}\, $$

$$ \begin{align} f'(x) & = \lim_{h \to 0}{(x+h)^n-x^n\over{h}} \\ & = \lim_{h \to 0}{(x^n + nx^{n-1}h + \cfrac{n(n-1)}{2!}x^{n-2}h^2 + \cdots + h^n)-x^n\over{h}} \\ & = \lim_{h \to 0}{nx^{n-1}h + \cfrac{n(n-1)}{2!}x^{n-2}h^2 + \cdots + h^n)\over{h}} \\ & = \lim_{h \to 0}{h(nx^{n-1} + \cfrac{n(n-1)}{2!}x^{n-2}h + \cdots + h^{n-1})\over{h}} \\ & = \lim_{h \to 0}{nx^{n-1} + \cfrac{n(n-1)}{2!}x^{n-2}h + \cdots + h^{n-1}} \\ & = \lim_{h \to 0}{nx^{n-1}} \\ & = nx^{n-1} \end{align} $$

Given:

$$ f(x) = c\, $$

Where c is a constant, prove:

$$ f'(x) = 0\, $$

$$ \begin{align} f'(x) & = \lim_{h \to 0}{f(x+h) - f(x)\over{h}} \\ & = \lim_{h \to 0}{c - c\over{h}} \\ & = \lim_{h \to 0}{0} \\ & = 0 \end{align} $$

Given:

$$ f(x)= mx + b\, $$

Prove:

$$ f'(x) = m\, $$

$$ \begin{align} f'(x) & = \lim_{h \to 0}{f(x+h) - f(x)\over{h}} \\ & = \lim_{h \to 0}{(m(x+h) + b) - (mx + b)\over{h}} \\ & = \lim_{h \to 0}{mx + mh + b - mx - b\over{h}} \\ & = \lim_{h \to 0}{mh\over{h}} \\ & = \lim_{h \to 0}{m} \\ & = m \end{align} $$