anonymous
  • anonymous
For f(x)=-2x^2 + 3x + 1 find (f(2+h)-f(2))/h Simplify completely and check your solution.
Algebra
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
f(2) and upon solving its answer is -1 so basically the end product is -2h-5
anonymous
  • anonymous
but here's the catch I don't know how to check this thing ?
Nnesha
  • Nnesha
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 \[\large\rm f(2+h) = -2\color{ReD}{(2+h)^2} +3\color{reD}{(2+h)}+1\]

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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{ \color{red}{f(h+2)}-\color{blue}{f(h)} }{h} }\) \(\large\color{black}{ \displaystyle \frac{ \color{red}{-2(h+2)^2+3(h+2)+1}-\color{blue}{-2h^2 + 3h + 1} }{h} }\)
SolomonZelman
  • SolomonZelman
Correction: \(\large\color{black}{ \displaystyle \frac{ \color{red}{-2(h+2)^2+3(h+2)+1}-\color{blue}({-2h^2 + 3h + 1)} }{h} }\)
anonymous
  • anonymous
I repeat it's f(h+2)-f(2)/h not f(h+2)-f(h)/h
Nnesha
  • Nnesha
it should be like this \[\large\rm \frac{( -2\color{Red}{(2+h)^2}+3\color{Red}{(2+h)}+1) - (2\color{ReD}{(2)}^2+3\color{ReD}{(2)}+1)}{ h }\]
anonymous
  • anonymous
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: $$D(x^2)=\frac{(h+2)^2-2^2}h=\frac{h^2+2h\color{red}{+4-4}}h=\frac{h^2+2h}h=h+2\\D(x)=\frac{(h+2)-2}h=\frac{h\color{red}{+2-2}}h=\frac{h}h=1\\D(1)=\frac{1-1}h=\frac0h=0$$
anonymous
  • anonymous
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.
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @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 \[\large\rm \frac{( -2\color{Red}{(2+h)^2}+3\color{Red}{(2+h)}+1) - (-2\color{ReD}{(2)}^2+3\color{ReD}{(2)}+1)}{ h }\]
anonymous
  • anonymous
oops, that should read \(h^2+4h\) on top which gives \(D(x^2)=h+4 \)
anonymous
  • anonymous
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.
anonymous
  • anonymous
now put it all together: so then it follows that $$\frac{f(h+2)-f(2)}h=D(-2x^2+3x+1)\\\qquad\qquad\qquad\quad=-2\cdot D(x^2)+3\cdot D(x)+D(1)\\\qquad\qquad\qquad\quad=-2(h+4)+3(1)+0\\\qquad\qquad\qquad\quad=-2h-8+3\qquad\qquad\qquad\qquad=\color{green}{-2h-5}$$
anonymous
  • anonymous
Correct but how to check this thing ? I'm done trying after one hour so that's why came here today.
anonymous
  • anonymous
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 \(f(3)-f(2)=-7\)
anonymous
  • anonymous
and then similarly with \(h=2\), maybe \(h=3\), etc.
anonymous
  • anonymous
okay and then what exactly does h=2 or h=3 check ? I mean the end result would be different as well.
anonymous
  • anonymous
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 \(h=2\)
anonymous
  • anonymous
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.
Nnesha
  • Nnesha
well i will check my work again .... hm
Nnesha
  • Nnesha
https://santosverdin63.wordpress.com/2012/02/01/54/
anonymous
  • anonymous
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.
anonymous
  • anonymous
*nnesha
anonymous
  • anonymous
@abs202 no, you are very confused, unfortunately, but i don't think i'm capable of helping correct that right now
UsukiDoll
  • UsukiDoll
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 \[\large -2h-5 \]
UsukiDoll
  • UsukiDoll
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.
UsukiDoll
  • UsukiDoll
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 \[f'(x) = -4x+3\] ??????????????
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
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) \[\large f'(x)=10x+3\]
ali2x2
  • ali2x2
^ correct

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