Question

Show that the function \( f(x) = 7x^2 - 3 \) is an increasing function when \( x>0 \).

Hint:

{Key Point:}

• Increasing function means \( f'(x)>0 \)
• For \( f(x) = 7x^2 - 3 \), \( f'(x) = 14x \)
• When \( x>0 \), \( 14x>0 \) → function is increasing

Answer 1

Step 1: Definition of an increasing function.
A function \(f(x)\) is said to be increasing on an interval if its derivative \(f'(x)\) is positive for all \(x\) in that interval. In other words, if \(f'(x) > 0\) for all \(x > 0\), then the function is increasing for \(x > 0\).

Step 2: Find the derivative of the function \(f(x)\).
The given function is: \[ f(x) = 7x^2 - 3 \] We differentiate \(f(x)\) with respect to \(x\): \[ f'(x) = \frac{d}{dx}(7x^2 - 3) = 14x \] Step 3: Analyze the derivative.
The derivative of the function is \( f'(x) = 14x \). For \( x > 0 \), we see that \( 14x > 0 \) because \( x \) is positive. Hence, \( f'(x) > 0 \) for all \( x > 0 \).

Step 4: Conclusion.
Since \( f'(x) > 0 \) for all \( x > 0 \), the function \( f(x) = 7x^2 - 3 \) is increasing when \( x > 0 \).

Final Answer: The function \( f(x) = 7x^2 - 3 \) is increasing for \( x > 0 \).