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

[karush]

$\tiny{APC.2.1.06}$
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:

 

[skeeter]

fyi, this is a calculator active problem …

 

[HOI]

$\tiny{APC.2.1.06}$
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!

 

[karush]

ok guess I was assuming that

fyi, this is a calculator active problem …

View attachment 11165

what graphing program is that from?
doesnt look like Desmos

 

[skeeter]

ok guess I was assuming that

what graphing program is that from?
doesnt look like Desmos

Good Grapher Pro app on my ipad.