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(note: my mathematical knowledge is around high school niveau. So please explain mathematical terms or use simple language - thank you :))

I have this task description:

  • $f(x) = e^{-x}$
  • tangent of $f(x)$ at $x = 0 \to t(x) = -x + 1$
  • $n(x) = x + 1$
  • my task is to prove that $t(x)$ and $n(x)$ form a isosceles triangle with the x-axis.

Question 1:

My own attempt to find an solution was rather complicated, but i still don't know if its even correct:

  • lets define the intersection point of $t(x)$ with the x-axis $P$ $\rightarrow (1,0)$
  • lets also define the intersection point of $t(x)$ with $n(x)$ as $M(0|1)$
  • We know that $n(x) \bot t(x)$
  • we rotate the triangle at $M(0|1)$ until $t(x) \bot$ x-axis. Let's call that new function $T(x)$. $T(x)$ is orthogonal to the y-axis.
  • we now have a rectangle $ABCD$
    • $A = $ the distance between $M(0|1)$ and the x-axis
    • $B = $ the distance between $O(0|0)$ and $P$
    • $C = $ the distance between $M(0|1)$ and $R(1|1)$
      • $T(x)$ goes straight thorugh both points
    • $D = $ the distance between $R(1|1)$ to $Z(0|1)$
    • its a rectangle because we know that $n(x) \bot T(x)$
  • by being a rectangle we know that the distance between $O(0|0)$ and $P$ has to be the same as the distance between $M(0|1)$ and $L(P_x|1)$ ($P_x$ is the x-value of $P$, which is $1$.)

Therefore we can conclude that also the distance from $n(0)$ to the x-axis has to be the same as from $t(0)$ to the x-axis. which in turn proves that the triangle is isosceles.

Is my argumentation mathematically correct?

Question 2:

My teacher's solution said that the angle between the x-axis and $n(x)$ and the angle between the x-axis and $t(x)$ has to be both $45^°$.

I haven't understood his solution, so basically, i want to know how i can prove the isosceles triangle with the $45^°$ desribed above. I want to highlight (again) that my mathematical knowledge is rather low, so please don't use complicated mathematical terms or explain them in easy terms.

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    I did not understand how you got rectangle. Can you elaborate bit more? Also what is $P_x$? – 2017-01-27
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    @VikrantDesai i provided some more information about the problem. Is it now more clear? $P_x$ is the x-value of the intersection point of $t(x)$ with the x-axis – 2017-01-27
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    I think you have over-complicated the problem. @Arnaldo has provided very easy to understand solution. I didn't see how you rotate a triangle to get a rectangle. When you rotate that triangle as you have said, the $t(x)$ actually becomes $y-$axis and $n(x)$ becomes a straight line parallel to $x-$axis and passing through the point $M[0|1]$. – 2017-01-27

1 Answers 1

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You can do something a bit easier:

$1)$ $t(x)$ and $n(x)$ meets at the point $A=(0,1)$

$2)$ $t(x)=-x+1$ meets the axis-x at $B=(1,0)$

$3)$ $n(x)=x+1$ meets the axis-x at $C=(-1,0)$

Now just calculate the sides:

$AB=\sqrt{(1-0)^2+(0-1)^2}=\sqrt{2}$

$AC=\sqrt{(-1-0)^2+(0-1)^2}=\sqrt{2}$

$BC=\sqrt{(-1-1)^2+(0-0)^2}=2$

So, $AB=AC$ and the triangle is isosceles.

P.S:

a) About the teacher's solution:

$t(x)=-x+1$ has angular coeficient equal to $-1$ and then if we call $\theta$ the inclination angle of the line then $\tan \theta=-1 \to \theta=180°-45°$ and then the internal angle of the triangle is $45°$.

$n(x)=x+1$ has angular coeficient equal to $1$ and then if we call $\alpha$ the inclination angle of the line then $\tan \alpha=1 \to \alpha=45°$. In this case $\alpha$ is already the internal angle.

See,

enter image description here

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    Thank you for this simple solution. But still I'd like to know how the argumentation with the $45^°$ goes and if my own solution is accurate. But still thank you. :) – 2017-01-27
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    @toogley: I updated about your teacher's solution. I'm taking a look in your solution – 2017-01-27
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    @toogley: Sorry but your solution is very confusing. First you confuse vertex and sides at the rectangle. Once $A$ is a vertex and then it is a distante. Furthermore I don't understand what is $T(x)$ and how you made your rotation. I recomend you take a look in my answer and in your teacher's answer, which are more simple. Leave rotations and constructions for a more tough problem. This is just a hint. – 2017-01-27