y = 1 π + r 2 y=\frac{1}{\pi+r^2} y = π + r 2 1
Domain: r ∈ ( − ∞ , ∞ ) r\isin(-\infin,\infin) r ∈ ( − ∞ , ∞ )
Asymptote:
lim r → − ∞ f ( r ) = lim r → ∞ f ( r ) = 0 \displaystyle\lim_{r\to -\infin}f(r)=\displaystyle\lim_{r\to \infin}f(r)=0 r → − ∞ lim f ( r ) = r → ∞ lim f ( r ) = 0
Horizontal asymptote is y = 0 y=0 y = 0
Symmetry:
there is symmetry respect to y-axis: f ( r ) = f ( − r ) f(r)=f(-r) f ( r ) = f ( − r )
Critical point, extrema:
y ′ = − 2 r ( π + r 2 ) 2 = 0 ⟹ r = 0 y'=-\frac{2r}{(\pi+r^2)^2}=0\implies r=0 y ′ = − ( π + r 2 ) 2 2 r = 0 ⟹ r = 0
the function is increasing on r ∈ ( − ∞ , 0 ) r\isin(-\infin,0) r ∈ ( − ∞ , 0 ) , y ′ > 0 y'>0 y ′ > 0
the function is decreasing on r ∈ ( 0 , ∞ ) r\isin(0,\infin) r ∈ ( 0 , ∞ ) , y ′ < 0 y'<0 y ′ < 0
maxima is ( 0 , 1 / π ) (0,1/\pi) ( 0 , 1/ π )
Inflection points:
y ′ ′ = − 2 ( π + r 2 ) 2 − 2 r ⋅ 4 r ( π + r 2 ) ( π + r 2 ) 4 = 0 y''=-\frac{2(\pi+r^2)^2-2r\cdot4r(\pi+r^2)}{(\pi+r^2)^4}=0 y ′′ = − ( π + r 2 ) 4 2 ( π + r 2 ) 2 − 2 r ⋅ 4 r ( π + r 2 ) = 0
4 r 2 − π − r 2 = 0 4r^2-\pi-r^2=0 4 r 2 − π − r 2 = 0
r = ± π / 3 r=\pm\sqrt{\pi/3} r = ± π /3
Inflection points are ( π / 3 , 1 π + 3 ) , ( − π / 3 , 1 π + 3 ) (\sqrt{\pi/3},\frac{1}{\pi+3}), (-\sqrt{\pi/3},\frac{1}{\pi+3}) ( π /3 , π + 3 1 ) , ( − π /3 , π + 3 1 )
Intercept point:
r = 0 ⟹ y = 1 / π r=0\implies y=1/\pi r = 0 ⟹ y = 1/ π
y-intercept point is ( 0 , 1 / π ) (0,1/\pi) ( 0 , 1/ π )
there no x-intercepts
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