Find the slope of the tangent line to the graph of f at the given point.
Take the derivative and plug in the x given. This is the DEFINITION of the slope of the tangent line at a point.
What is the derivative?
Have you learned the `Power Rule for Derivatives` yet?
No I have not.
Have you learned about the `Limit Definition of the Derivative` ?\[\Large f'(x) \quad=\quad \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]
I have a basic understanding of it.
The derivative, when evaluated at a particular point, gives us the slope of the line tangent to the function at that point. I know I know, that's a bit of a mouthful. Let's just... yah. So we have,\[\Large f(x)=x^2+5x\]And we want to evaluate the derivative function at x=4. The y coordinate doesn't really do anything for us. We just need the 4.\[\Large f'(4) \quad=\quad \lim_{h\to0}\frac{f(4+h)-f(4)}{h}\]
OK
So it looks like for this setup, we're going to need to calculate f(4) and f(4+h).
Understand how to evaluate the function at a particular point? Here is how f(4) would look just in case there's any confusion:\[\Large f(\color{royalblue}{x})=(\color{royalblue}{x})^2+5(\color{royalblue}{x})\]Plugging 4 in gives us,\[\Large f(\color{royalblue}{4})=(\color{royalblue}{4})^2+5(\color{royalblue}{4})\]
I got f(4)=36
And that reduces to 9, so I reckon I'd pick C.
What reduces to 9? :o
Well I thought I'd divide 36 by 4..........
f(4) is a special function notation. There is no 4 to be divided by D: It's saying, plug 4 in for x and then simplify. So we've established so far that,\[\Large f(4)=36\]Ok good.
\[\Large f(\color{royalblue}{x})=(\color{royalblue}{x})^2+5(\color{royalblue}{x})\]We also need to find this part\[\Large f(\color{royalblue}{4+h})=(\color{royalblue}{4+h})^2+5(\color{royalblue}{4+h})\],
I don't know how to work that out.
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