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anonymous

  • 5 years ago

Explain local linear approximation and how to do it please?

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  1. anonymous
    • 5 years ago
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    Of one variable or two? Are you in Cal I?

  2. anonymous
    • 5 years ago
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    I think it's 1 variable. Here's the question: For the function f, f ' (x)=2x+1 and f(1) = 4. What is the approximation for f(1.2) found by using the line tangent to the graph of f at x=1? I'm in AP Calc AB

  3. anonymous
    • 5 years ago
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    Because it is a AP question, there is a lot of trickery involved rather than a straightforward question. You can start of by finding f(x). I imagine that would be the integral of f'(x)=2x+1

  4. anonymous
    • 5 years ago
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    I imagine f(1)=4 should have been f'(1)=4?

  5. anonymous
    • 5 years ago
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    no it just says f(1) = 4. I got x^2 + x + C as f(x) through integration. I remember you're supposed to do local linear approximation to find C, but I don't remember how & I can't find my notes from that day.

  6. anonymous
    • 5 years ago
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    Local linear approximation would be f(1)+f'(1)(1)(1.2-1)

  7. anonymous
    • 5 years ago
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    Decoding that: evaluate f at one, add to (f' evaluated at one, multiplied by 1.2-1). I accidentally wrote an extra (1) above.

  8. anonymous
    • 5 years ago
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    So 4.6? Thanks so much.

  9. anonymous
    • 5 years ago
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    I think there is a clue in there find C. f(1)=4. so therefore (1^2) +1 + C=4 Solve for C

  10. anonymous
    • 5 years ago
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    Check your answer. Redo everything with the value f(x)=x^2+x+2 (Now that we know the value of C.

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