Question

For what value of c will y= x^2 + (c/x) have a relative minimum at x=-1?
a: c=-4
b: c=-2
c: c=2
d: c=4
e: none of these

Answer 1

\[y=x^2+\frac{ c }{ x }\]
\[\frac{ dy }{ dx }=2x-\frac{ c }{ x^2 }\]
it has a relative minimum at x=-1
Hence \[\frac{ dy }{ dx }=0~at~x=-1\]
\[2*-1+\frac{ c }{(- 1)^2 }=0,c=2\]

Answer 2

Thanks!

Answer 3

yw