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# If f(x) is a quadratic expression such that f(x)>0 ∀ x ∈ R, and if g(x)=f(x)+f'(x)+f''(x), then prove that g(x) > 0 ∀ x ∈ R.

Jitender Singh IIT Delhi
7 years ago
Ans:
Hello Student,

Let

where a, b & c belong to real
Since f(x) > 0 for all x
a > 0 &
Discriminent < 0
b2< 4ac..............(1)



Let find discriminent



So discriminent is -ve.
So g(x) does not have any real root.
g(x) > 0 for all x belong to real.

mycroft holmes
272 Points
7 years ago
We see that g(x) is a quadratic with the same leading coefficient as f(x) and hence g(x) has a minimum. Let that occur at x = x0.

Then at xo, we have

$g'(x_0) = f'(x_0)+f"(x_0) =0 \ (\text{note that} f"'(x)=0)$

Then we have

$g(x_0) = f(x_0)+ f'(x_0)+f"(x_0) = f(x_0)>0 \ (\because f(x)>0 \ \forall x \in \mathbb{R})$

Since the minimum value of g(x) is positive, we have that g(x) has no real roots.