Question

If f(x) is quadratic expression which is positive for all real value of x and g(x)=f(x)+f′(x)+f′′(x)
. Then for any real value of x

Answer 1

knowing that f(x) > 0 for all x does not guarantee that f'(x) > 0.
Consider f(x) = x^2+1
f'(x) = 2x is negative for all x < 0.

Answer 2

We know that f(x) is a quadratic expression that is positive for all real values of x. This means that the graph of f(x) is a parabola that opens upwards. Therefore, the vertex of the parabola is at the minimum point, and the minimum value of f(x) is greater than 0.

Now, let's consider g(x) = f(x) + f'(x) + f''(x). We know that f(x) is positive for all x, so f'(x) is also positive for all x. This means that f'(x) is increasing for all x, and f''(x) is positive for all x.

Since f''(x) is positive for all x, the sum of f(x), f'(x), and f''(x) will also be positive for all x. This means that g(x) is positive for all real values of x.

In conclusion, for any real value of x, g(x) will always be positive.

Answer 3

You are correct, I apologize for the mistake in my previous response.

If we consider f(x) = x^2 + 1 as an example, we have:

f'(x) = 2x

f''(x) = 2

Now let's calculate g(x) = f(x) + f'(x) + f''(x):

g(x) = (x^2 + 1) + (2x) + 2
g(x) = x^2 + 2x + 3

Since g(x) is a quadratic expression, we can complete the square to find its minimum point. The equation can be rewritten as:

g(x) = (x + 1)^2 + 2

Since (x + 1)^2 is always non-negative, the minimum value of g(x) is 2, which means g(x) is always greater than or equal to 2 for any real value of x.

Therefore, for any real value of x, g(x) will always be greater than or equal to 2.