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: Recall the condition for increasing function A function \( f(x) \) is said to be increasing in an interval if: \[ f'(x)>0 \quad \text{for all } x \text{ in that interval} \]
Step 2: Find the derivative of \( f(x) \) \[ f(x) = 7x^2 - 3 \] \[ f'(x) = \frac{d}{dx}(7x^2 - 3) = 14x \]
Step 3: Check the sign of \( f'(x) \) for \( x>0 \) For \( x>0 \): \[ f'(x) = 14x>0 \] Since \( 14x \) is positive for all positive values of \( x \), we have: \[ f'(x)>0 \quad \forall \, x>0 \]
Step 4: Conclusion Since \( f'(x)>0 \) for all \( x>0 \), the function \( f(x) = 7x^2 - 3 \) is an increasing function when \( x>0 \). \[ \boxed{\text{The function is increasing for } x>0} \] Graphical Interpretation: The function \( f(x) = 7x^2 - 3 \) is a parabola opening upwards with vertex at \( (0, -3) \). For \( x>0 \), as \( x \) increases, \( f(x) \) increases continuously.