A differentiable function `f(x)` has a relative minimum at `x=0.` Then the function `f=f(x)+a x+b` has a relative minimum at `x=0` for (a)all `a` and all`b` (b) all `b` if `a=0` (c)all `b >0` (d) all `a >0`

Options

A. all a and all b
B. all b if a =0
C. all b `gt` 0
D. all a `gt` 0

Answer

2

Explanation

Since f(x) has a relative minimum at x =0 f(X)=0 and `f(0)gt0` (assuming `f(X)ne 0)`
If the funciton y =f(X) ax+b has a relative minimum at x=0 then
`(dy)/(dx)=0 at x =0 or f(X) + a =0 for x =0`
F(0)+a=0 or a =0
or a=0
Now `(d^(2)y)/(dx^(2))=f''(x)or (d^(2)y)/(dx^(2))_(x=0)=f''(0)gt0 [therefore f''(0)gt0]`
Hence y has a relative minimum at x= 0 if a =0 and b can attain anyreal value.