Given the function f(x) = x^4 - 5x + 2 . Which of the following is the slope when x=2?
a. 22
b. 7
c. 28
d. -28
PLEASE HELP

- anonymous

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- DebbieG

Do you know how to take the derivative of f(x)?

- DebbieG

You just need to take the derivative, then plug x=2 into the derivative. that gives you the slope of the graph at x=2.

- anonymous

@DebbieG hello when i plugged x into the equation i got an answer of 4 which isnt one of the choices so im messing up somewhere

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## More answers

- DebbieG

What did you plug x=2 into? You said "the equation"... you aren't supposed to plug x=2 into f(x) (that just gives you f(2)). (I'm not sure if that's what you did though, because f(2) is not =4).
You need to first take the derivative of the f(x). Did you do that? What did you get?

- DebbieG

\[f\prime(x) = \frac{ d }{ dx } (x^4 - 5x + 2)\]

- DebbieG

Given that you are expected to know how to do this problem, I'm assuming that you've covered derivatives, at least for this type of function?

- anonymous

@DebbieG well this is all new to me we received homework after a class session where all this was introduced including derivatives

- DebbieG

Hey, wait a second... the correct answer isn't there. Are you sure you've given the function correctly?

- DebbieG

is the middle term -5x^2, by chance?

- anonymous

yes its x^4 -5x + 2 thats the whole equation

- DebbieG

It's fine that it's new... you have to start somewhere. :) But it's important that you understand the rules for taking the derivative of a polynomial before you can tackle this problem.

- DebbieG

Well then the answer is not there, but if instead, the function is
\(f(x)=x^4 - 5x^2 + 2\)
then the answer IS there. So I think that's a typo, and we should proceed with that assumption.

- anonymous

@DebbieG okay well can you help me through the steps of solving it please so i can know exactly how the answer was developed

- DebbieG

Now to differentiate any term of a polynomial, e.g., anything that looks like \(5x^8\) or \(-2x^4\) or \(x^2\) --- anything where you have a constant and then a power of x - the rule is simple:
Pull the power out in front, multiply it by the constant (note that the constant might be 1), and then subtract one from the power.
E.g.:
\(\Large \frac{ d }{ dx } 3x^2=2\cdot3x^{2-1}=6x\)
\(\Large \frac{ d }{ dx } (-5x^4)=4\cdot-5x^{4-1}=-20x^3\)
\(\Large \frac{ d }{ dx } (x^6)=6\cdot x^{6-1}=6x^5\)
Now look those over, do you understand the rule? Do you see how I computed each derivative?

- DebbieG

Also notice:
\(\Large \frac{ d }{ dx } (2x)=2\cdot x^{1-1}=2x^0=2\) (since \(b^0=1\))
So in other words, the derivative of a constant times x is just whatever the constant is.
And the last rule you'll need is this:
\(\Large \frac{ d }{ dx } c=0\) for any constant \(\large c\), e.g., the derivative of a constant is = 0.

- DebbieG

Also, for polynomials, you get the derivative of the whole polynomial by just adding up the derivatives of each term... e.g.:
\(\Large f\prime(x) = \frac{ d }{ dx } (x^4 - 5x^2 + 2)= \frac{ d }{ dx } (x^4)+ \frac{ d }{ dx } (- 5x^2)+ \frac{ d }{ dx } (2)\)
Just take each derivative and then add them up. Remember, we are assuming that middle term should be an \(x^2\) term.

- DebbieG

Now can you put that all together and tell me what our \(\Large f\prime(x)\) is? E.g., what do you get from:
\(\Large \frac{ d }{ dx } (x^4)+ \frac{ d }{ dx } (- 5x^2)+ \frac{ d }{ dx } (2)\)

- anonymous

@debbieG is f'(X)=7 ?

- DebbieG

No. First of all, f'(x) here will be a FUNCTION OF x. Not a constant. (The only time it would be a constant is if you just have a linear expression, so the only x in the function is a 1st degree term, e.g., f(x)=7x+1. Then by the differentiation rules I gave you above, f'(x)=7.
Can you tell me, step by step, what you did? I want to see where you went wrong. Did you understand the rules I gave you above?

- DebbieG

We want
\(\Large \frac{ d }{ dx } (x^4)+ \frac{ d }{ dx } (- 5x^2)+ \frac{ d }{ dx } (2)\)
So start with the first piece: what is \(\Large \frac{ d }{ dx } (x^4)\)? Just read up above, I told you exactly how to do it and showed you several examples.

- DebbieG

It's most similiar to my last example, where I did:
\(\Large \frac{ d }{ dx } (x^6)=6\cdot x^{6-1}=6x^5\)

- DebbieG

Now you do \(\Large \frac{ d }{ dx } (x^4)\)

- anonymous

|dw:1378225846418:dw|

- anonymous

@DebbieG is that the correct shown above

- DebbieG

Right, so that's the derivative of the first term.
\(\Large \frac{ d }{ dx } (x^4)=4x^3\)
Now try the second term, same idea:
\(\Large \frac{ d }{ dx } (- 5x^2)=??\)

- anonymous

|dw:1378228055050:dw|

- anonymous

@DebbieG see above answer

- DebbieG

Very good. Now you just have the last term to do, and then we'll put it all together:
\(\Large \frac{ d }{ dx } (2)=??\)
Remember what I said up above, about the derivative of a constant (this is the easy one! :)

- anonymous

@DebbieG is it 0

- goformit100

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- DebbieG

RIGHT! Now put it all together:
\(\Large f\prime(x) = \frac{ d }{ dx } (x^4 - 5x^2 + 2)= \frac{ d }{ dx } (x^4)+ \frac{ d }{ dx } (- 5x^2)+ \frac{ d }{ dx } (2)=???\)
What is your function, f'(x)?

- anonymous

@DebbieG 22

- DebbieG

Hmmm..... I think there is still something wrong here with the function. I was thinking that f'(2) would give us 22 (so I thought you had it!!), but I see now that I was wrong.
You see why I think there is a typo in the function? If that middle term isn't an \(x^2\) term then you don't get any of your answer choices for f'(2).
But one step at a time, using my "modification" of the function first you had to cook up f'(x):
\(\Large f\prime(x) = \frac{ d }{ dx } (x^4 - 5x^2 + 2)= 4x^3-10x\), right? That's what we did, step by step, one term at a time.
Then to get f'(2) you just plopped the 2 in for the x:
\(\Large f\prime(2) = 4\cdot2^3-10\cdot2=4\cdot8-20=32-20=12\)
That was my mistake.... I did it in my head earlier, and thought that would give us 22. My bad!
But given your original function, we get almost the same derivative, except that middle term would have a derivative of -5, instead of -5x:
\(\Large f\prime(x) = \frac{ d }{ dx } (x^4 - 5x + 2)= 4x^3-5\)
So then we get:
\(\Large f\prime(2) = 4\cdot2^3-5=32-5=27\)
..... also not one of the choices (unless b was supposed to say "27"??)

- DebbieG

Do you see what I'm saying? I think there's a typo somewhere here... either in the problem as given to you or in what you posted. The slope when x=2 for the function you actually posted is f'(2)=27.

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