Suppose that `f(x)` isa quadratic expresson positive for all real `xdot` If `g(x)=f(x)+f^(prime)(x)+f^(x),` then for any real `x(w h e r ef^(prime)(x)a n df^(x)` represent 1st and 2nd derivative, respectively). `g(x)<0` b. `g(x)>0` c. `g(x)=0` d. `g(x)geq0`

Options

A. `g(x) lt 0`
B. `g(x) gt 0`
C. `g(x) = 0`
D. `g(x) ge 0`

Answer

2

Explanation

Let `f(x) = ax^(2) + bx + c` be a quadratic expression such that
`f(x) gt 0` for all `x in R`. Then, `a gt 0` and `b^(2) - 4ac lt 0`. Now,
`g(x) = f(x) + f'(x) + f''(x)`
`rArr g(x) = ax^(2) + x (b + 2a) + (b + 2a + c)`
Discriminant of g(x) is
`D = (b + 2a)^(2) - 4a (b + 2a + c)`
`= b^(2) - 4a^(2) - 4ac`
`= (b^(2) - 4ac) - 4a^(2)`
`lt 0" " (because b^(2) - 4ac lt 0)`
Therefore, `g(x) gt 0` fro all `x in R`.