Question

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

Answer 1

f(2) and upon solving its answer is -1
so basically the end product is -2h-5

Answer 2

but here's the catch I don't know how to check this thing ?

Answer 3

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 \[\large\rm f(2+h) = -2\color{ReD}{(2+h)^2} +3\color{reD}{(2+h)}+1\]

Answer 4

\(\large\color{black}{ \displaystyle \frac{ \color{red}{f(h+2)}-\color{blue}{f(h)} }{h} }\)

\(\large\color{black}{ \displaystyle \frac{ \color{red}{-2(h+2)^2+3(h+2)+1}-\color{blue}{-2h^2 + 3h + 1} }{h} }\)

Answer 5

Correction:
\(\large\color{black}{ \displaystyle \frac{ \color{red}{-2(h+2)^2+3(h+2)+1}-\color{blue}({-2h^2 + 3h + 1)} }{h} }\)

Answer 6

I repeat it's f(h+2)-f(2)/h not f(h+2)-f(h)/h

Answer 7

it should be like this \[\large\rm \frac{( -2\color{Red}{(2+h)^2}+3\color{Red}{(2+h)}+1) - (2\color{ReD}{(2)}^2+3\color{ReD}{(2)}+1)}{ h }\]

Answer 8

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: $$D(x^2)=\frac{(h+2)^2-2^2}h=\frac{h^2+2h\color{red}{+4-4}}h=\frac{h^2+2h}h=h+2\\D(x)=\frac{(h+2)-2}h=\frac{h\color{red}{+2-2}}h=\frac{h}h=1\\D(1)=\frac{1-1}h=\frac0h=0$$

Answer 9

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.

Answer 10

\(\color{blue}{\text{Originally Posted by}}\) @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 \[\large\rm \frac{( -2\color{Red}{(2+h)^2}+3\color{Red}{(2+h)}+1) - (-2\color{ReD}{(2)}^2+3\color{ReD}{(2)}+1)}{ h }\]

Answer 11

oops, that should read \(h^2+4h\) on top which gives \(D(x^2)=h+4 \)

Answer 12

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.

Answer 13

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

Answer 14

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

Answer 15

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 \(f(3)-f(2)=-7\)

Answer 16

and then similarly with \(h=2\), maybe \(h=3\), etc.

Answer 17

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

Answer 18

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 \(h=2\)

Answer 19

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.

Answer 20

well i will check my work again .... hm

Answer 21

https://santosverdin63.wordpress.com/2012/02/01/54/

Answer 22

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.

Answer 23

*nnesha

Answer 24

@abs202 no, you are very confused, unfortunately, but i don't think i'm capable of helping correct that right now

Answer 25

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 \[\large -2h-5 \]

Answer 26

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.

Answer 27

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
\[f'(x) = -4x+3\]
??????????????

Answer 28

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

Answer 29

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)
\[\large f'(x)=10x+3\]

Answer 30

^ correct