Heyyy....I’m still confused. Nah jk but thx for the points lol
Take derivitive
note
the derivitive of sec(x)=sec(x)tan(x)
so
remember the quotient rule
the derivitive of

so
the derivitive of
so now evaluate when t=pi
we get
sec(pi)=-1
tan(pi)=0
we get

slope=1/pi
use slope point form
for
slope=m and point is (x1,y1)
equation is
y-y1=m(x-x1)
slope is 1/pi
point is (pi,1/pi)
y-1/π=1/π(x-π)
times both sides by π
πy-1=x-π
πy=x-π+1
y=(1/π)x-1+(1/π)
or, alternately
-(1/π)x+y=(1/π)-1
x-πy=π-1
X = 5 or x = 13.
If you need details, please ask!
Find the absolute value vertex. In this case, the vertex for y=|x−5|y=|x-5| is (5,0)(5,0).
(5,0)(5,0)
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
(−∞,∞)(-∞,∞)
{x|x∈R}{x|x∈ℝ}
For each xx value, there is one yy value. Select few xx values from the domain. It would be more useful to select the values so that they are around the xx value of the absolute valuevertex.
xy3241506172