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Average change rate of a function #21: f(x)=1/2+3
- Thread starter Illvoices
- Start date
Well, first off I'm guessing there's a small typo here and the given function is meant to be f(x) = (1/2)x+3. What you've posted is the horizontal line y = 7/2. That aside, what have you tried? Please share with us any and all work you've done on this problem, even if you know it's wrong. If you need a refresher on this material, you might try here.
Okay, so what I think your (poorly formatted) working is showing is this:
\(\displaystyle \dfrac{\dfrac{1}{2}x+3}{100-100h} = \dfrac{\dfrac{1}{2}(100)+3 - \dfrac{1}{2}(100h)+3}{h}\)
Assuming the above is the correct interpretation of your work, I have a few questions. First, where did the number 100 come from? The problem you originally posted asked about the average rate of change between the points x=a and x=a+h. It appears as if you've somehow transformed that into the interval between the points x=100 and x=100h. By what basis have you concluded this? Second, by what basis have you concluded that the denominator, being 100-100h, is equal to h?
\(\displaystyle \dfrac{\dfrac{1}{2}x+3}{100-100h} = \dfrac{\dfrac{1}{2}(100)+3 - \dfrac{1}{2}(100h)+3}{h}\)
Assuming the above is the correct interpretation of your work, I have a few questions. First, where did the number 100 come from? The problem you originally posted asked about the average rate of change between the points x=a and x=a+h. It appears as if you've somehow transformed that into the interval between the points x=100 and x=100h. By what basis have you concluded this? Second, by what basis have you concluded that the denominator, being 100-100h, is equal to h?
Okay, so what I think your (poorly formatted) working is showing is this:
\(\displaystyle \dfrac{\dfrac{1}{2}x+3}{100-100h} = \dfrac{\dfrac{1}{2}(100)+3 - \dfrac{1}{2}(100h)+3}{h}\)
Assuming the above is the correct interpretation of your work, I have a few questions. First, where did the number 100 come from? The problem you originally posted asked about the average rate of change between the points x=a and x=a+h. It appears as if you've somehow transformed that into the interval between the points x=100 and x=100h. By what basis have you concluded this? Second, by what basis have you concluded that the denominator, being 100-100h, is equal to h?
actually the h at the end was on top and i would like to know how'd you get to write the equation that good with the fraction line and all
This forum has a built in LaTeX parser, to facilitate greater of ease of "writing in math," as it were. Using it is fairly simple. For instance, if you wanted to write a faction, you would simply write in your message:
which would then produce:
\(\displaystyle \dfrac{numerator}{denominator}\)
You can "nest" fractions as well by placing one inside another. The code for the left-hand side of the equation in my previous post was made this code:
\(\displaystyle \dfrac{\dfrac{1}{2}x+3}{100-100h}\)
You can find a decent guide on LaTeX here.
LaTeX can be a helpful tool, but it's not really going to immediately help you solve the problem you were tasked with. I asked a few questions in my last post to clear up exactly what your reasoning was, so as to know how best to proceed. Did you see those questions?
Code:
[tex]\dfrac{numerator}{denominator}[/tex]which would then produce:
\(\displaystyle \dfrac{numerator}{denominator}\)
You can "nest" fractions as well by placing one inside another. The code for the left-hand side of the equation in my previous post was made this code:
Code:
[tex]\dfrac{\dfrac{1}{2}x+3}{100-100h}[/tex]\(\displaystyle \dfrac{\dfrac{1}{2}x+3}{100-100h}\)
You can find a decent guide on LaTeX here.
LaTeX can be a helpful tool, but it's not really going to immediately help you solve the problem you were tasked with. I asked a few questions in my last post to clear up exactly what your reasoning was, so as to know how best to proceed. Did you see those questions?