Prove that the function given by
$
f(x) = x^3 - 3x^2 + 3x = 100
$
is increasing on $ \mathbb{R} $.

Q: Prove that the function given by $ f(x) = x^3 - 3x^2 + 3x = 100 $ is increasing on $ \mathbb{R} $.

Explanation: Let \[ f(x) = x^3 - 3x^2 + 3x \] Differentiate $ f(x) $ with respect to $ x $: \[ f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 3x) = 3x^2 - 6x + 3 \] Factor the derivative: \[ f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x - 1)^2 \] Now, for all $ x \in \mathbb{R} $, we know: \[ (x - 1)^2 \geq 0 \Rightarrow f'(x) = 3(x - 1)^2 \geq 0 \] Also, $ f'(x) = 0 $ only at $ x = 1 $, and $ f'(x) > 0 $ elsewhere. Thus, $ f'(x) \geq 0 $ for all $ x \in \mathbb{R} $, and $ f'(x) > 0 $ for all $ x \ne 1 $. \[\] $\textbf{Conclusion:}$ $ f(x) = x^3 - 3x^2 + 3x \text{ is an increasing function on } \mathbb{R}. $

Question

Prove that the function given by  
$
f(x) = x^3 - 3x^2 + 3x = 100
$  
is increasing on $ \mathbb{R} $.

Accepted Answer

Explanation: Let  
$$
f(x) = x^3 - 3x^2 + 3x
$$

Differentiate $ f(x) $ with respect to $ x $:  
$$
f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 3x) = 3x^2 - 6x + 3
$$

Factor the derivative:  
$$
f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x - 1)^2
$$

Now, for all $ x \in \mathbb{R} $, we know:  
$$
(x - 1)^2 \geq 0 \Rightarrow f'(x) = 3(x - 1)^2 \geq 0
$$

Also, $ f'(x) = 0 $ only at $ x = 1 $, and $ f'(x) > 0 $ elsewhere.  
Thus, $ f'(x) \geq 0 $ for all $ x \in \mathbb{R} $, and $ f'(x) > 0 $ for all $ x \ne 1 $.

$\textbf{Conclusion:}$  
$
f(x) = x^3 - 3x^2 + 3x \text{ is an increasing function on } \mathbb{R}.
$

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Question:

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