Results 1 to 2 of 2

Math Help - Average rate of change

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    83

    Average rate of change

    I am unsure as to which forum to put this in. I took a shot and put it here, apologies if wrong.

    Okay, to the question I am having trouble with.

    "Find the average rate of change of f(x) = \sqrt{x + 11} with respect to x from x = 5 to x = 5 +h. [4 marks]"

    My attempt to answer:
    f(5) = \sqrt{5 + 11}
    f(5) = \sqrt{16}
    f(5) = 4

    f(5 + h) = \sqrt{5 + h + 11}
    f(5 + h) = \sqrt{16 + h}
    f(5 + h) = 4 + \sqrt{h}
    *I am unsure if I am correct in square rooting the 16, so having 4 + the square root of h.*

    Then I use the \frac{y2 - y1}{x2 - x1} formula to get the slope of the secant *rate of change*:

    \frac{4 + \sqrt{h} - 4}{5 - 5 + h}
    = \frac{\sqrt{h}}{h}
    That's my final solution. My problem with this solution is when I sub a number in for h, I get a different answer from running the numbers through, and simply going *square root of the number divided by the number*.
    Example:
    Let's pretend h is 20.
    So I have f(5) = 4.
    I can sub in 20 for h:
    f(5 + 20) = \sqrt{5 + 20 + 11}
    f(25) = \sqrt{36}
    f(25) = 6

    Run that through the slope formula:
    \frac{6 - 4}{20 - 5}
    = \frac{2}{5}
    = 0.4

    Then, using my average rate of change thing: \frac{\sqrt{h}}{h} I get:
    \frac{\sqrt{20}}{20}
    = 0.223606797
    which is a completely different answer. So I must be doing something wrong here.

    Okay, I tried using 25 for h in the \frac{\sqrt{h}}{h} and it gives the same ansewr as running the math through. So it is not technically h, but it is h + 5, so is this actually right??? I see that as: \frac{\sqrt{h+5}}{h+5} which I think is different, please correct me, I am so confused.

    EDIT:There, I have successfully turned this into something readable using latex
    Last edited by Kakariki; September 15th 2009 at 01:04 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by Kakariki View Post
    I am unsure as to which forum to put this in. I took a shot and put it here, apologies if wrong.

    Okay, to the question I am having trouble with.

    "Find the average rate of change of f(x) = \sqrt{x + 11} with respect to x from x = 5 to x = 5 +h. [4 marks]"

    My attempt to answer:
    f(5) = \sqrt{5 + 11}
    f(5) = \sqrt{16}
    f(5) = 4

    f(5 + h) = \sqrt{5 + h + 11}
    f(5 + h) = \sqrt{16 + h}
    f(5 + h) = 4 + \sqrt{h}
    *I am unsure if I am correct in square rooting the 16, so having 4 + the square root of h.*

    Then I use the \frac{y2 - y1}{x2 - x1} formula to get the slope of the secant *rate of change*:

    \frac{4 + \sqrt{h} - 4}{5 - 5 + h}
    = \frac{\sqrt{h}}{h}
    That's my final solution. My problem with this solution is when I sub a number in for h, I get a different answer from running the numbers through, and simply going *square root of the number divided by the number*.
    Example:
    Let's pretend h is 20.
    So I have f(5) = 4.
    I can sub in 20 for h:
    f(5 + 20) = \sqrt{5 + 20 + 11}
    f(25) = \sqrt{36}
    f(25) = 6

    Run that through the slope formula:
    \frac{6 - 4}{20 - 5}
    = \frac{2}{5}
    = 0.4

    Then, using my average rate of change thing: \frac{\sqrt[h]][h]] I get:
    \frac{\sqrt{20}}{20}
    = 0.223606797
    which is a completely different answer. So I must be doing something wrong here.

    Okay, I tried using 25 for h in the \frac{\sqrt{h}}{h} and it gives the same ansewr as running the math through. So it is not technically h, but it is h + 5, so is this actually right???

    the wrong thing as what you said you can't separate 16 from the square root

    f(5+x)=\sqrt{16+h} you can't simplify it more than that


    s=\frac{\sqrt{16+h} - 4 }{h}\

    EDIT:There, I have successfully turned this into something readable using latex
    CONGRATULATIONS LATEX is simple and useful
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Average rate of change!
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: January 27th 2010, 07:32 PM
  2. Average Rate of Change
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 28th 2009, 01:54 PM
  3. average rate of change
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: September 27th 2009, 09:22 AM
  4. Average Rate of Change
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 27th 2008, 08:39 PM
  5. average rate of change
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 16th 2008, 09:55 AM

Search Tags


/mathhelpforum @mathhelpforum