1.

Show that the function `f(x)=x^2+3x+5`,is continuos at x=1.

Answer» Atx = 1
`underset(xrarr1)R.H.L= underset(hrarr0)limf(x)=limf(1+h)`
=`lim{(1+h)^2+3(1+h)+5}`
`=(1+0)^(2)+3(1+0)+5=9`
`L.H.L =underset(xrarr1)limf(x)=underset(hrarr0)limf(1-h)`
`underset(hrarr0)lim{(1-h)^(2)+3(1-h)+5}`
`=(1-0)^2+3(1-0)+5=9`
and f(1)`=1^2+3(1)+5=9`
Now R.H.L = f(1)=L.H.L
`therefore f(x)=x^2+3x+5` is continuous at x=1


Discussion

No Comment Found

Related InterviewSolutions