Question #155402

Sketch the graph of

y=1/π+r2

by finding the domain, symmetries, critical points, inflection points,

intercept points, asymptotes, extremas, intervals on which the function is increasing or decreasing,

concave up or down.




Expert's answer

y=1π+r2y=\frac{1}{\pi+r^2}


Domain: r∈(−∞,∞)r\isin(-\infin,\infin)


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

Horizontal asymptote is y=0y=0


Symmetry:

there is symmetry respect to y-axis: f(r)=f(−r)f(r)=f(-r)


Critical point, extrema:

y′=−2r(π+r2)2=0  ⟹  r=0y'=-\frac{2r}{(\pi+r^2)^2}=0\implies r=0

the function is increasing on r∈(−∞,0)r\isin(-\infin,0) , y′>0y'>0

the function is decreasing on r∈(0,∞)r\isin(0,\infin) , y′<0y'<0

maxima is (0,1/π)(0,1/\pi)


Inflection points:

y′′=−2(π+r2)2−2r⋅4r(π+r2)(π+r2)4=0y''=-\frac{2(\pi+r^2)^2-2r\cdot4r(\pi+r^2)}{(\pi+r^2)^4}=0

4r2−π−r2=04r^2-\pi-r^2=0

r=±π/3r=\pm\sqrt{\pi/3}

Inflection points are (π/3,1π+3),(−π/3,1π+3)(\sqrt{\pi/3},\frac{1}{\pi+3}), (-\sqrt{\pi/3},\frac{1}{\pi+3})


Intercept point:

r=0  ⟹  y=1/πr=0\implies y=1/\pi

y-intercept point is (0,1/π)(0,1/\pi)

there no x-intercepts

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