Use the limit process to find the slope of the graph of
sqrt(x + 8) at (8, 4).
Here's a similar problem:
http://www.jiskha.com/display.cgi?id=1415653817
Post again if you have further questions.
This is a very different problem.
This deals with the slope not the area
It's almost the same problem, slope is the same thing as the area between the region. But use the same procedures..
Integrate the function.
Then plug in x=4 and x=8
f(4)-f(8)
So is there an answer or is it undefined
To find the slope of the graph of the function $\sqrt{x+8}$ at a specific point (8,4), you can use the limit process. The slope of the function at a given point is equivalent to the derivative of the function at that point.
The derivative of a function can be found using the following limit definition:
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
$$
In this case, let's find the derivative (slope) of the function $\sqrt{x+8}$ at the point (8,4).
Substitute the given point (8,4) into the derivative expression:
$$
f'(8) = \lim_{h \to 0} \frac{f(8+h)-f(8)}{h}
$$
Next, substitute the function $f(x) = \sqrt{x+8}$ into the expression above:
$$
f'(8) = \lim_{h \to 0} \frac{\sqrt{8+h+8} - \sqrt{8+8}}{h}
$$
Simplify the expression further:
$$
f'(8) = \lim_{h \to 0} \frac{\sqrt{h+16} - 4}{h}
$$
To evaluate this limit, we can multiply both the numerator and denominator of the fraction by the conjugate of the numerator, which is $\sqrt{h+16} + 4$:
$$
f'(8) = \lim_{h \to 0} \frac{\sqrt{h+16}-4}{h} \times \frac{\sqrt{h+16} + 4}{\sqrt{h+16} + 4}
$$
After simplifying, we get:
$$
f'(8) = \lim_{h \to 0} \frac{(h+16) - 16}{h(\sqrt{h+16} + 4)}
$$
Now we can cancel out the terms and evaluate the limit:
$$
f'(8) = \lim_{h \to 0} \frac{h}{h(\sqrt{h+16} + 4)}
$$
$$
f'(8) = \lim_{h \to 0} \frac{1}{\sqrt{h+16} + 4}
$$
Finally, substitute h = 0 into the expression:
$$
f'(8) = \frac{1}{\sqrt{0+16} + 4} = \frac{1}{4+4} = \frac{1}{8}
$$
Therefore, the slope of the graph of $\sqrt{x+8}$ at the point (8,4) is 1/8.