Let f(x) be a quadratic expression possible for all real x.
If `g(x)=f(x)-f'(x)+f''(x),` then for any real x

Options

A. `g(x)gt0`
B. `g(x)le0`
C. `g(x)ge0`
D. `g(x)lt0`

Answer

To solve the problem, we need to analyze the function \( g(x) = f(x) - f'(x) + f''(x) \), where \( f(x) \) is a quadratic function. Let's break it down step by step.

### Step 1: Define the quadratic function
Let \( f(x) = ax^2 + bx + c \), where \( a, b, c \) are constants and \( a > 0 \) for the quadratic to open upwards.

**Hint:** Remember that a quadratic function can be expressed in the standard form \( ax^2 + bx + c \).

### Step 2: Calculate the first derivative \( f'(x) \)
The first derivative of \( f(x) \) is:
\[ f'(x) = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b \]

**Hint:** The derivative of \( ax^n \) is \( nax^{n-1} \).

### Step 3: Calculate the second derivative \( f''(x) \)
The second derivative of \( f(x) \) is:
\[ f''(x) = \frac{d}{dx}(2ax + b) = 2a \]

**Hint:** The derivative of a linear function \( mx + b \) is simply \( m \).

### Step 4: Substitute \( f(x) \), \( f'(x) \), and \( f''(x) \) into \( g(x) \)
Now, substitute these derivatives into the expression for \( g(x) \):
\[ g(x) = f(x) - f'(x) + f''(x) \]
\[ g(x) = (ax^2 + bx + c) - (2ax + b) + 2a \]

### Step 5: Simplify \( g(x) \)
Now, simplify the expression:
\[ g(x) = ax^2 + bx + c - 2ax - b + 2a \]
\[ g(x) = ax^2 + (b - 2a)x + (c - b + 2a) \]

**Hint:** Combine like terms carefully.

### Step 6: Analyze the discriminant of \( g(x) \)
The discriminant \( D \) of a quadratic \( Ax^2 + Bx + C \) is given by \( D = B^2 - 4AC \). For \( g(x) \):
- \( A = a \)
- \( B = b - 2a \)
- \( C = c - b + 2a \)

Thus, the discriminant \( D \) is:
\[ D = (b - 2a)^2 - 4a(c - b + 2a) \]

### Step 7: Simplify the discriminant
Now, we need to simplify this expression:
\[ D = (b - 2a)^2 - 4a(c - b + 2a) \]
\[ D = b^2 - 4ab + 4a^2 - 4ac + 4ab - 8a^2 \]
\[ D = b^2 - 4ac - 4a^2 \]

### Step 8: Determine the sign of \( D \)
Since we know that \( b^2 - 4ac < 0 \) (as given in the problem), we have:
\[ D = (b^2 - 4ac) - 4a^2 < 0 - 4a^2 < 0 \]

This means \( D < 0 \).

### Step 9: Conclusion about \( g(x) \)
Since the discriminant of \( g(x) \) is less than 0, \( g(x) \) has no real roots and opens upwards (as \( a > 0 \)). Therefore, \( g(x) > 0 \) for all real \( x \).

**Final Answer:** \( g(x) > 0 \) for all real \( x \).

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