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Math Help - I have no idea what this is called or how to solve

  1. #1
    dto
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    I have no idea what this is called or how to solve

    Please help

    Using the given function, find f(x+h) - f(x) / h

    A.) f(x) = 3x-7

    B.) F(X)=X^2 +3X -2


    I would show work, but I have no idea what to do... I know the answers I just really want to know how to solve.

    Thanks
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  2. #2
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    Okay. So the problem is asking you to substitute various expressions for f(x) into the formula

    \frac{f(x+h)-f(x)}{h}

    So all you do is take the expression and plug it in where f(x) is. For example, for the first one the expression is 3x-7. So we want to first replace any f(x) we find with 3x-7:

    \frac{f(x+h)-(3x-7)}{h}

    Second, we see that f(x+h) is also a function, even though it doesn't look exactly like f(x). What we have to do to figure out f(x+h) is do the exact same thing we did with the regular f(x) - we substituted 3x-7 - except instead of just plain x, we do it with x+h. So:

    f(x+h) = 3(x+h)-7

    Now, since we've figured out f(x+h), we can substitute the value we got back into the original equation. Now we have:

    \frac{3(x+h)-7 -(3x-7)}{h}

    When we multiply this out and simplify algebraically, we get

    \frac{3h}{h} = 3

    What this is is really a beginning calculus problem; it's the formula for finding a derivative using limits. So you might get better answers trying the calculus board. Can you solve the second one on your own? If not, and if I didn't explain in enough detail, just ask and I'll try to help you out some more.
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  3. #3
    Flow Master
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    Quote Originally Posted by dto View Post
    Please help

    Using the given function, find f(x+h) - f(x) / h

    A.) f(x) = 3x-7

    B.) F(X)=X^2 +3X -2


    I would show work, but I have no idea what to do... I know the answers I just really want to know how to solve.

    Thanks
    It's a prelude to when you will be studying calculus (finding the derivative of the function f(x) from first principles).
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  4. #4
    dto
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    Awesome Awsome! Thanks alot, cleared it all up.

    So would the answer for part B, be 2x+h+3?
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    No problem! I'm glad you understood, cause I wasn't entirely sure I made sense. And yes, 2x+h+3 is totally right. Nice work!

    It may be interesting to note that in most calculus problems, instead of the exact value of

    \frac{f(x+h)-f(x)}{h},

    they would probably ask you to find the limit as h tends to zero:

    \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}

    In this case, to find the limit you would just calculate the value of the expression like before, and then you would evaluate it when h=0. So,

    \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h} when f(x) = x^2+3x-2

    is simply

    \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h} = 2x+h+3

    and, just plugging in 0 for h, we get

    \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h} = 2x+3
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