-apc.2.1.06 crosses the x-axis at one point in the interval [0,1]

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Discussion Overview

The discussion revolves around the problem of determining the slope of the graph of the function $y=e^{\tan x}-2$ at the point where it crosses the x-axis within the interval [0,1]. Participants explore the mathematical steps involved in finding this slope, including the use of logarithms and derivatives.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant outlines the steps to find the x-coordinate where the graph crosses the x-axis, leading to the conclusion that $x=\arctan(\ln 2) \approx 0.606$.
  • The same participant calculates the slope at this point using the derivative formula $y_m=e^{\tan (x)}\sec^2(x)$ and arrives at a slope of approximately 2.96.
  • Another participant points out a potential misunderstanding in the formulation of the problem, suggesting that the initial equation should be stated as $e^{\tan(x)}-2=0$ for clarity.
  • There is a mention of the problem being a "calculator active problem," implying reliance on computational tools for solving it.
  • Participants discuss the graphing program used, with one participant mentioning the "Good Grapher Pro" app on their iPad.

Areas of Agreement / Disagreement

There is no clear consensus on the correctness of the initial mathematical formulation, as one participant questions the clarity of the problem statement. The discussion remains unresolved regarding the best approach to solving the problem.

Contextual Notes

Some assumptions about the problem formulation and the use of computational tools are present, but these are not fully explored or clarified.

karush
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$\tiny{APC.2.1.06}$
The graph of $y=e^{\tan x}-2$ crosses the x-axis at one point in the interval [0,1]
What is the slope of the graph at this point.
a, 0.606 b. 2 c, 2.242 d.2.961 e.3.747

[d]
$\begin{array}{rll}
\textit{given} &e^{\tan \:x}-2 &(1)\\
\textit{rewrite} &e^{\tan x}=2 &(2) \\
\textit{e thru} &\tan x=\ln 2 &(3)\\
\textit{isolate x } &x=\arctan \ln 2+\pi n &(4)\\
\textit{[0,1] } &\arctan(\ln 2)=0.606 &(5)\\
\textit{m at x} &y_m=e^{\tan (x)}\sec ^2(x) &(6)\\
\textit{m} &y_m(0.606)=2.96 &(7)
\end{array}$

it took me 2 hours to do this
must be a better way

if its correct:unsure:
 
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fyi, this is a calculator active problem …

6695B933-7821-4557-9A00-DAC241A97F13.jpeg
 
karush said:
$\tiny{APC.2.1.06}$
The graph of $y=e^{\tan x}-2$ crosses the x-axis at one point in the interval [0,1]
What is the slope of the graph at this point.
a, 0.606 b. 2 c, 2.242 d.2.961 e.3.747

[d]
$\begin{array}{rll}
\textit{given} &e^{\tan \:x}-2 &(1)\\
\textit{rewrite} &e^{\tan x}=2 &(2) \\
\textit{e thru} &\tan x=\ln 2 &(3)\\
\textit{isolate x } &x=\arctan \ln 2+\pi n &(4)\\
\textit{[0,1] } &\arctan(\ln 2)=0.606 &(5)\\
\textit{m at x} &y_m=e^{\tan (x)}\sec ^2(x) &(6)\\
\textit{m} &y_m(0.606)=2.96 &(7)
\end{array}$

it took me 2 hours to do this
must be a better way

if its correct:unsure:
You mean "given $e^{\tan(x)}-2= 0$". Otherwise your second line makes no sense!
 
ok guess I was assuming that
skeeter said:
fyi, this is a calculator active problem …

View attachment 11165
what graphing program is that from?
doesnt look like Desmos
 
karush said:
ok guess I was assuming that

what graphing program is that from?
doesnt look like Desmos

Good Grapher Pro app on my ipad.
 

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