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To solve for the given question, we need to find the derivative of the function $f(x) = 3x^2 – 4x + 1$.The derivative of a function $f(x)$ with respect to $x$, denoted as $f'(x)$, represents the rate of change of $f(x)$ with respect to $x$. To find the derivative of $f(x) = 3x^2 – 4x + 1$, we need to apply the power rule for differentiation. The power rule states that if $f(x) = x^n$, where $n$ is a real number, then $f'(x) = nx^{n-1}$.Let us apply this rule term by term for $f(x) = 3x^2 – 4x + 1$.1. For the term $3x^2$: The power rule applied to $x^2$ gives us $2x^{2-1} = 2x$. However, we need to keep the constant coefficient $3$ in place. Thus, the derivative of $3x^2$ is: $$ \frac{d}{dx}(3x^2) = 3 \cdot \frac{d}{dx}(x^2) = 3 \cdot 2x = 6x. $$2. For the term $-4x$: The power rule applied to $x$ (which is $x^1$) gives us $1x^{1-1} = 1 \cdot x^0 = 1$ (since $x^0 = 1$ for $x \neq 0$). However, we need to keep the constant coefficient $-4$ in place. Thus, the derivative of $-4x$ is: $$ \frac{d}{dx}(-4x) = -4 \cdot \frac{d}{dx}(x) = -4 \cdot 1 = -4. $$3. For the constant term $1$: The derivative of any constant term is zero. Thus, the derivative of $1$ is: $$ \frac{d}{dx}(1) = 0. $$4. Combining all the terms together, the derivative $f'(x)$ is: $$ f'(x) = 6x – 4 + 0 = 6x – 4. $$So, the derivative of $f(x) = 3x^2 – 4x + 1$ is: $$ \boxed{6x – 4}. $$To verify, let us recall a few more details about the power rule explicitly for a term $ax^n$ where $a$ is a constant: $$ \frac{d}{dx}(ax^n) = a \cdot nx^{n-1}. $$ This makes it easier to directly apply the rule to each term individually.### Verification by First Principles (if needed for a deeper understanding)The derivative can also be defined as: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}. $$ For $f(x) = 3x^2 – 4x + 1$: $$ f(x+h) = 3(x+h)^2 – 4(x+h) + 1. $$ Expanding $3(x+h)^2$: $$ 3(x+h)^2 = 3(x^2 + 2xh + h^2) = 3x^2 + 6xh + 3h^2. $$ So, $$ f(x+h) = 3x^2 + 6xh + 3h^2 – 4(x+h) + 1 = 3x^2 + 6xh + 3h^2 – 4x – 4h + 1. $$ We need to find $f(x+h) – f(x)$: $$ f(x+h) – f(x) = (3x^2 + 6xh + 3h^2 – 4x – 4h + 1) – (3x^2 – 4x + 1). $$ $$ = 3x^2 + 6xh + 3h^2 – 4x – 4h + 1 – 3x^2 + 4x – 1. $$ $$ = 3x^2 – 3x^2 + 6xh + 3h^2 – 4x + 4x – 4h + 1 – 1. $$ $$ = 6xh + 3h^2 – 4h. $$ $$ = h(6x + 3h – 4). $$ Now, substitute this back into the limit definition of the derivative: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h} = \lim_{h \to 0} \frac{h(6x + 3h – 4)}{h}. $$ $$ = \lim_{h \to 0} (6x + 3h – 4). $$ $$ = 6x + 3(0) – 4. $$ $$ = 6x – 4. $$ This confirms our result once again.### Conclusion The derivative of $f(x) = 3x^2 – 4x + 1$ is: $$ \boxed{6x – 4}. $$ \boxed{6x – 4}\boxed{“Fix4Bot.com Aims to Repair and Support Nvidia’s ‘Supercharged’ Humanoid Robot Developments”}
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