Let's evaluate and simplify the given expressions step by step for the function $f(x) = 6x^2 - x + 7$.

1. Finding $f(x + h)$:

We are asked to find $f(x+h)$, which means we substitute $x + h$ in place of $x$ in the original function.

The original function is: $f(x) = 6x^2 - x + 7$

Now, substitute $x + h$ into the function: $f(x + h) = 6(x + h)^2 - (x + h) + 7$

Next, expand the terms: $(x + h)^2 = x^2 + 2xh + h^2$ $f(x + h) = 6(x^2 + 2xh + h^2) - (x + h) + 7$

Now, distribute the 6: $f(x + h) = 6x^2 + 12xh + 6h^2 - x - h + 7$

This is the fully simplified expression for $f(x+h)$: $f(x + h) = 6x^2 + 12xh + 6h^2 - x - h + 7$

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2. Finding $\frac{f(x + h) - f(x)}{h}$:

We now want to find the expression for $\frac{f(x+h) - f(x)}{h}$.

We already know: $f(x + h) = 6x^2 + 12xh + 6h^2 - x - h + 7$ $f(x) = 6x^2 - x + 7$

Now, subtract $f(x)$ from $f(x + h)$: $f(x + h) - f(x) = (6x^2 + 12xh + 6h^2 - x - h + 7) - (6x^2 - x + 7)$

Distribute the negative sign: $f(x + h) - f(x) = 6x^2 + 12xh + 6h^2 - x - h + 7 - 6x^2 + x - 7$

Simplify the terms: $f(x + h) - f(x) = 12xh + 6h^2 - h$

Now, divide the entire expression by $h$: $\frac{f(x + h) - f(x)}{h} = \frac{12xh + 6h^2 - h}{h}$

Factor out the $h$ from the numerator: $\frac{f(x + h) - f(x)}{h} = \frac{h(12x + 6h - 1)}{h}$

Cancel the $h$ in the numerator and denominator (assuming $h \neq 0$): $\frac{f(x + h) - f(x)}{h} = 12x + 6h - 1$

This is the simplified expression for $\frac{f(x + h) - f(x)}{h}$.

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Final Answers:

1. $f(x + h) = 6x^2 + 12xh + 6h^2 - x - h + 7$
2. $\frac{f(x + h) - f(x)}{h} = 12x + 6h - 1$

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Would you like further details on any of these steps? Let me know if you have any questions!

Here are some related questions to expand your understanding:

1. How do these expressions relate to the concept of the derivative?
2. What happens to $\frac{f(x + h) - f(x)}{h}$ as $h$ approaches 0?
3. How would you apply this to find the derivative of a function?
4. What is the significance of the term $12x + 6h - 1$ when $h \to 0$?
5. How can you use the concept of limits to evaluate derivatives?

Tip: When simplifying expressions like this, always be careful with distributing and combining like terms.