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QUESTION: I could really benefit from some help with an old question of mine (got it wrong on homework a long while ago- was sick the day this material was taught- never quite got the question).

I drew up a graph of my question on a Paint program: http://prntscr.com/5s9w4d

The graph of f' is shown in the link. If we know that f(2)=10, then the local linearization of f at x=2 is f(x)=?

My theory was A, but I am really not sure on it. I I could benefit from some help with an old question of mine (got it wrong on homework a long while ago- was sick the day this material was taught- never quite got the question). Thank you, Clark

ANSWER: Clark, I couldn't open the link. Please resend. Not sure if you attached it properly.

[an error occurred while processing this directive]---------- FOLLOW-UP ----------

QUESTION: http:// prntscr .com/ 5s9w4d

Try the above web address (not an attachment; type it into the address bar, but without the spaces)

Got the plot. If the curve is the derivative of f(x), then it can be modeled as a line near x = 2

f'(x) = mx + b = (1/2)x + 2

Thus, f(x) is the integral given by

f(x) = (1/4)x^2 + 2x + c.

To evaluate the constant c, we know f(2) = 10 = (1/4)(2^2) + 2(2) + c or

c = 5.

Now, for the linearization, I assume this means to use just the 1st 2 terms in the Taylor series of f(x) about x = 2, which translates into

f_lin(x-2) ≈ f(2) + f'(2)･(x-2) = 10 + 3(-2) = 4 + 3x

where I have used f'(2) = 3.

Looking back on the problem, the formula for f_lin is probably something they just gave you in class without using the "Taylor series" approach, so your task would have probably just to have taken f(2)= 10 and read off f'(2) = 3 and plugged it in, without bothering to derive the equation for f(x). In any case, there you have it.

BTW, I'm not sure what you meant by "my theory was A". Please send me a follow-up if I've misread your question.

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