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Prove the following.
Prove that the function f(x) = x³ - 6x² + 12x + 5 is increasing on mathbbR.
Prove that the function $f(x) = x^3 - 6x^2 + 12x + 5$ is increasing on $\mathbb{R}$.
Given: $f(x) = x^3 - 6x^2 + 12x + 5$
To show: $f(x)$ is increasing on $\mathbb{R}$
Answer
Answer
$f'(x) \geq 0$ for all real $x$, so $f$ is increasing on $\mathbb{R}$.
- 1. $f(x) = x^3 - 6x^2 + 12x + 5$, differentiable on $\mathbb{R}$.
- 2. $f'(x) = 3x^2 - 12x + 12$ (power rule).
- 3. $f'(x) = 3(x^2 - 4x + 4) = 3(x-2)^2$ (factorising).
- 4. $(x-2)^2 \geq 0$ for every real $x$, so $f'(x) \geq 0$ for all $x \in \mathbb{R}$, with $f'(x) = 0$ only at $x = 2$ (square is non-negative).
- 5. Since $f'(x) \geq 0$ on $\mathbb{R}$ and vanishes at an isolated point only, $f$ is increasing on $\mathbb{R}$ (first derivative test for monotonicity).