We are given the function $f(x) = 2x^2$ and asked to find the simplified form of the difference quotient, and then complete a table.

(a) Simplified form of the difference quotient:

The difference quotient is given by the formula:

$\frac{f(x+h) - f(x)}{h}$

Now, let's calculate $f(x+h)$. Since $f(x) = 2x^2$, we have:

$f(x+h) = 2(x+h)^2$

Now substitute this expression and $f(x) = 2x^2$ into the difference quotient formula:

$\frac{f(x+h) - f(x)}{h} = \frac{2(x+h)^2 - 2x^2}{h}$

Simplify $(x+h)^2$ first:

$(x+h)^2 = x^2 + 2xh + h^2$

So,

$f(x+h) = 2(x^2 + 2xh + h^2) = 2x^2 + 4xh + 2h^2$

Now, substitute $f(x+h)$ and $f(x) = 2x^2$ into the difference quotient:

$\frac{2x^2 + 4xh + 2h^2 - 2x^2}{h}$

Cancel out the $2x^2$ terms:

$\frac{4xh + 2h^2}{h}$

Now factor out $h$ from the numerator:

$\frac{h(4x + 2h)}{h}$

Cancel out the $h$'s:

$4x + 2h$

Thus, the simplified form of the difference quotient is:

$\frac{f(x+h) - f(x)}{h} = 4x + 2h$

(b) Completing the table:

Now we can use the simplified form of the difference quotient to complete the table. The difference quotient formula is:

$4x + 2h$

For $x = 4$, we substitute $x = 4$ into the expression $4x + 2h$ for each value of $h$:

• When $h = 2$:

$4(4) + 2(2) = 16 + 4 = 20$

• When $h = 1$:

$4(4) + 2(1) = 16 + 2 = 18$

• When $h = 0.1$:

$4(4) + 2(0.1) = 16 + 0.2 = 16.2$

• When $h = 0.01$:

$4(4) + 2(0.01) = 16 + 0.02 = 16.02$

Completed table:

$x$	$h$	$\frac{f(x+h) - f(x)}{h}$
4	2	20
4	1	18
4	0.1	16.2
4	0.01	16.02

Do you want a further explanation of any part of this solution? Here are some related questions you can explore:

1. How does the difference quotient relate to the derivative of a function?
2. What would happen if we had a more complex function, like $f(x) = x^3$?
3. How does the value of $h$ affect the accuracy of the difference quotient in approximating the derivative?
4. What is the significance of the derivative in real-world applications?
5. How would you find the derivative using the limit definition as $h \to 0$?

Tip: As $h$ gets smaller, the difference quotient gets closer to the derivative of the function at that point!