Question

Show that the function \( f(x) = x^3 + 10x + 7 \), \( x \in \mathbb{R} \), is strictly increasing.

Hint:

A function is strictly increasing if its derivative is positive for all values of \( x \).

Answer 1

To show that a function is strictly increasing, we need to prove that its derivative is positive for all values of \( x \). Step 1: Find the derivative of the function \( f(x) = x^3 + 10x + 7 \): \[ f'(x) = 3x^2 + 10 \] Step 2: Check the sign of \( f'(x) \): \[ f'(x) = 3x^2 + 10 \] Since \( 3x^2 + 10 \) is always positive (because \( x^2 \geq 0 \) for all \( x \) and the constant 10 is positive), we conclude that \( f'(x)>0 \) for all \( x \in \mathbb{R} \). Step 3: Conclusion: Since the derivative is always positive, the function \( f(x) = x^3 + 10x + 7 \) is strictly increasing for all values of \( x \in \mathbb{R} \).