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anonymous

  • one year ago

For f(x)=-2x^2 + 3x + 1 find (f(2+h)-f(2))/h Simplify completely and check your solution.

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  1. anonymous
    • one year ago
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    f(2) and upon solving its answer is -1 so basically the end product is -2h-5

  2. anonymous
    • one year ago
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    but here's the catch I don't know how to check this thing ?

  3. Nnesha
    • one year ago
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    f(2+h) means x = 2+h so you need to find f(2+h) replace x by 2+h in \(\color{red}{\rm f(x)}\) function \[\large\rm f(2+h) = -2\color{ReD}{(2+h)^2} +3\color{reD}{(2+h)}+1\]

  4. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{ \color{red}{f(h+2)}-\color{blue}{f(h)} }{h} }\) \(\large\color{black}{ \displaystyle \frac{ \color{red}{-2(h+2)^2+3(h+2)+1}-\color{blue}{-2h^2 + 3h + 1} }{h} }\)

  5. SolomonZelman
    • one year ago
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    Correction: \(\large\color{black}{ \displaystyle \frac{ \color{red}{-2(h+2)^2+3(h+2)+1}-\color{blue}({-2h^2 + 3h + 1)} }{h} }\)

  6. anonymous
    • one year ago
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    I repeat it's f(h+2)-f(2)/h not f(h+2)-f(h)/h

  7. Nnesha
    • one year ago
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    it should be like this \[\large\rm \frac{( -2\color{Red}{(2+h)^2}+3\color{Red}{(2+h)}+1) - (2\color{ReD}{(2)}^2+3\color{ReD}{(2)}+1)}{ h }\]

  8. anonymous
    • one year ago
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    break it into pieces: the difference quotient $$D(af+bg)=aD(f)+bD(g)$$ so let's look at how it beahves on \(x^2,x,1\) independently: $$D(x^2)=\frac{(h+2)^2-2^2}h=\frac{h^2+2h\color{red}{+4-4}}h=\frac{h^2+2h}h=h+2\\D(x)=\frac{(h+2)-2}h=\frac{h\color{red}{+2-2}}h=\frac{h}h=1\\D(1)=\frac{1-1}h=\frac0h=0$$

  9. anonymous
    • one year ago
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    yes correct nnesha but the thing is upon solving I got the final answer as (-2h-5) but could someone tell me how to check this thing ? It's given in the chapter of functions.

  10. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha it should be like this \[\large\rm \frac{( -2\color{Red}{(2+h)^2}+3\color{Red}{(2+h)}+1) - (2\color{ReD}{(2)}^2+3\color{ReD}{(2)}+1)}{ h }\] \(\color{blue}{\text{End of Quote}}\) correction it should be like this \[\large\rm \frac{( -2\color{Red}{(2+h)^2}+3\color{Red}{(2+h)}+1) - (-2\color{ReD}{(2)}^2+3\color{ReD}{(2)}+1)}{ h }\]

  11. anonymous
    • one year ago
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    oops, that should read \(h^2+4h\) on top which gives \(D(x^2)=h+4 \)

  12. anonymous
    • one year ago
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    I appreciate the help Oldrin but we have not covered the "difference quotient" topic yet but again as aforementioned this question is from the chapter of functions.

  13. anonymous
    • one year ago
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    now put it all together: so then it follows that $$\frac{f(h+2)-f(2)}h=D(-2x^2+3x+1)\\\qquad\qquad\qquad\quad=-2\cdot D(x^2)+3\cdot D(x)+D(1)\\\qquad\qquad\qquad\quad=-2(h+4)+3(1)+0\\\qquad\qquad\qquad\quad=-2h-8+3\qquad\qquad\qquad\qquad=\color{green}{-2h-5}$$

  14. anonymous
    • one year ago
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    Correct but how to check this thing ? I'm done trying after one hour so that's why came here today.

  15. anonymous
    • one year ago
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    to check, you could plug in values of \(h\), say, \(h=1\): $$\frac{f(1+2)-f(2)}1=f(3)-f(2)=\ ?=-2(1)-5=-7$$and then check to make sure \(f(3)-f(2)=-7\)

  16. anonymous
    • one year ago
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    and then similarly with \(h=2\), maybe \(h=3\), etc.

  17. anonymous
    • one year ago
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    okay and then what exactly does h=2 or h=3 check ? I mean the end result would be different as well.

  18. anonymous
    • one year ago
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    well, you would check that: $$\frac{f(2+2)-f(2)}2=\frac{f(4)-f(2)}2=\,? =-2(2)-5=-9$$ so you'd verify that \(\frac12(f(4)-f(2))=-9\) for \(h=2\)

  19. anonymous
    • one year ago
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    Oldrin, the answer which you gave is entirely different than the main question. The tricky part with checking the solution is you have to tie it back to the main equation. What you told is giving a whole new answer entirely.

  20. Nnesha
    • one year ago
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    well i will check my work again .... hm

  21. Nnesha
    • one year ago
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    https://santosverdin63.wordpress.com/2012/02/01/54/

  22. anonymous
    • one year ago
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    Net could you pls explain me that in the form of equations here, the thing is I'm not used to paragraph cum equation type of explanations.

  23. anonymous
    • one year ago
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    *nnesha

  24. anonymous
    • one year ago
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    @abs202 no, you are very confused, unfortunately, but i don't think i'm capable of helping correct that right now

  25. UsukiDoll
    • one year ago
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    For f(x)=-2x^2 + 3x + 1 find (f(2+h)-f(2))/h \[f(x) = -2x^2+3x+1\] we have to use the definition of the derivative which is \[\large \frac{f(x+h)-f(x)}{h}\] so our x = 2 for \[\large \frac{f(2+h)-f(2)}{h}\] If we just have x = 2 in the function we will just have \[f(2) = -2(2)^2+3(2)+1 =-2(4)+6+1=-8+6+1=-2+1=-1\] so that portion is correct now we use the definition of the derivative to solve \[f(x) = -2x^2+3x+1\] \[\large \frac{-2(2+h)^2+3(2+h)+1-(-1)}{h}\] \[\large \frac{-2(h^2+4h+4)+6+3h+1+1)}{h}\] \[\large \frac{-2h^2-8h-8+6+3h+1+1}{h}\] rearrange the terms \[\large \frac{-2h^2-8h+3h-8+6+1+1}{h}\] \[\large \frac{-2h^2-8h+3h}{h}\] \[\large \frac{-2h^2-5h}{h}\] Now we can factor an h and cancel it. \[\large \frac{(h)[-2h-5]}{h}\] so our final answer is \[\large -2h-5 \]

  26. UsukiDoll
    • one year ago
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    The only thing I can think of to make sure that you have -2h-5 is by doing the process all over again and if you have the same answer, then the arithmetic was done correctly. If there are different answers, then there was an error.

  27. UsukiDoll
    • one year ago
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    There is another method though. In Calculus I, the definition of the derivative is taught first before all the fancy shortcut derivatives come into play. Using the definition of the derivative and those derivative techniques should yield the same answer, but the problem is that the derivative of \[f(x) = -2x^2+3x+1 \] is \[f'(x) = -4x+3\] ??????????????

  28. UsukiDoll
    • one year ago
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    I'll provide an example sort of like a proof that derivative techniques and the definition of the derivative yield the same result. Consider the function \[\large f(x)=5x^2+3x-7\] The derivative is \[\large f'(x)=10x+3\] now compare this to the attachment

  29. UsukiDoll
    • one year ago
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    Consider the function \[\large f(x)=5x^2+3x-7\] since -7 is a constant, the derivative is 0 The derivative is \[\large f(x)=5(2)x^{2-1}+(1)3x^{1-1}+)\] \[\large f(x)=10x^{1}+3x^{0}\] \[\large x^0 = 1 \] (by zero exponent rule) \[\large f'(x)=10x+3\]

  30. ali2x2
    • one year ago
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    ^ correct

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