let f(x)=x2+1/x2 and g(x)=x-1/x , x belongs to R-{-1,0,1}, if h(x)=f(x)/g(x), then the minimum value of h(x) is - Maths - Limits and Derivatives

Question

let f(x)=x2+1/x2 and g​(x)=x-1/x , x
belongs to R-{-1,0,1}, if h​(x)=f(x)/g(x), then the minimum
value of h(x) is

Aarushi Mishra · Accepted Answer

fx=x2+1x2fx=x2+1x2-2+2fx=x2+1x2-2×x×1x+2fx=x-1x2+2gx=x-1xhx=fxgx=x-1x2+2x-1xLet x-1x=thx=pt=t2+2tpt=t+2tdptdt=1-2t2dptdt=t2-2t2For maxima or minima dptdt=0t2-2t2=0t2-2=0t2=2t=±2dptdt=1-2t2d2ptdt2=2×2t3At t=2,d2ptdt2=2×223>0minima at t=2At t=-2,d2ptdt2=2×223<0maxima at t=-2min hx=min  pt=p2=22+22=42=22

let f(x)=x2+1/x2 and g​(x)=x-1/x , x belongs to R-{-1,0,1}, if h​(x)=f(x)/g(x), then the minimum value of h(x) is

let f(x)=x²+1/x² and g(x)=x-1/x , x belongs to R-{-1,0,1}, if h(x)=f(x)/g(x), then the minimum value of h(x) is