Question

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

Hint:

To determine if a function is increasing or decreasing, always check the sign of its first derivative. A positive derivative means the function is increasing, a negative derivative means it's decreasing, and a zero derivative indicates a potential local maximum, minimum, or point of inflection.

Answer 1

Step 1: Understanding the Concept:
A function is considered increasing on an interval if its first derivative is positive on that interval. So, to show that \( f(x) \) is increasing for \( x>0 \), we need to find its derivative, \( f'(x) \), and show that \( f'(x)>0 \) for all \( x>0 \).
Step 2: Key Formula or Approach:
The condition for an increasing function is:
\[ f'(x)>0 \] The power rule for differentiation states that \( \frac{d}{dx}(x^n) = nx^{n-1} \).
Step 3: Detailed Explanation or Calculation:
First, we find the function given:
\[ f(x) = 7x^2 - 3 \] Next, we differentiate \( f(x) \) with respect to \( x \) to find \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(7x^2 - 3) \] \[ f'(x) = 7 \cdot \frac{d}{dx}(x^2) - \frac{d}{dx}(3) \] \[ f'(x) = 7(2x) - 0 \] \[ f'(x) = 14x \] Now, we need to check the condition for the function to be increasing, which is \( f'(x)>0 \).
We are given the condition that \( x>0 \).
If \( x>0 \), then multiplying by a positive constant (14) will not change the inequality sign.
\[ 14x>14(0) \] \[ 14x>0 \] So, \( f'(x)>0 \) for all \( x>0 \).
Step 4: Final Answer:
Since the first derivative \( f'(x) = 14x \) is positive for all \( x>0 \), the function \( f(x) = 7x^2 - 3 \) is an increasing function when \( x>0 \).