I have no idea how to solve this question.

a) Find the slope of the tangent line to the curve at the point $(1, 0)$.

The answer is -6.

Yet, how did they arrive at this answer using this formula:


Now the steps are shown as so:

(a) Using Definition 1: $m=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ with $f(x)=x-x^\color{red}7$ and $P(\color{red}1,\color{red}0)$,

$$\begin{align} m&=\lim_{x\to1}\frac{f(x)-\color{red}0}{x-\color{red}1}=\lim_{x\to1}\frac{x-x^\color{red}7}{x-\color{red}1}=\lim_{x\to1}\frac{x(1-x^\color{red}6)}{x-\color{red}1}\\&=\lim_{x\to1}\frac{x(1-x)\color{red}{(1+x+x^2+x^3+x^3+x^4+x^5)}}{x-\color{red}1}\\&=\lim_{x\to1}\left[-x\color{red}{\left(1+x+x^2+x^3+x^3+x^4+x^5\right)}\right]=-1(\color{red}6)=\color{red}{-6}. \end{align}$$

(b) An equation of the tangent line is $$\begin{align} &y-f(a)=f'(a)(x-a)\\\implies&y-f(\color{red}1)=f'(\color{red}1)(x-\color{red}1)\\\implies&y-\color{red}0=\color{red}{-6}(x-\color{red}1),\text{ or }y=-\color{red}{-6}x+\color{red}6 \end{align}$$

…but I don't understand them. Can someone explain these simply?

  • $\begingroup$ i think answer should be -6 . $\endgroup$ – avz2611 Feb 22 '15 at 16:21
  • 1
    $\begingroup$ Can you be more clear about exactly which of these steps you don't understand? $\endgroup$ – Gregory Grant Feb 22 '15 at 16:21
  • $\begingroup$ The second line 4th equation. How was that expanded? $\endgroup$ – Cetshwayo Feb 22 '15 at 16:27
  • $\begingroup$ $1-x^6 \text{ was factored } \\ \text{ You could divide } -x^6+1 \text{ by} -x+1 \text{ to find that other factor if you don't know the formula that is }$ $\endgroup$ – randomgirl Feb 22 '15 at 16:31

From a differential coefficient meaning / viewpoint slope is demonstrated to be -6. All that was discredited when you saw a new result given as 6.

You must see which is stronger evidence as you understood. A simple printer's devil/typo or the calculation of derivative from basics? Such self judgment is important all the time in maths.


Take the derivative of $f(x)=x-x^7$ which is $f'(x)=1-7x^6$ and evaluate it in $x_0=1$: you'll obtain $f'(1)=1-7=-6$ which is the slope of the tangent line to the curve $y=x-x^7$ at the point $(1,0)$. I think there must be a typo in your book.

  • $\begingroup$ They never explained within the book how to solve a problem such as this. What is the algorithm (strategy) called to solving problems such as this? Is it taking the derivative? $\endgroup$ – Cetshwayo Feb 22 '15 at 17:36
  • $\begingroup$ In general it's dangerous to speak about "algorithm" in Mathematics. Think at integration: there isn't a unique way to solve them, you have to experience yourself. However in this case you have to know some basic matters in calculus. The question you asked was based on the fact that the definition of derivative of a function $f$ in a given point $x_0$ is the angular coefficient of the tangent line to the graphic of the curve defined from your function $f$ at the given point $(x_0,f(x_0))$. $\endgroup$ – Joe Feb 22 '15 at 18:16

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