f(x) problems

Princezz3286

Junior Member
Joined
Nov 12, 2005
Messages
66
find and simplify for:
f(x)= -5x+2

f(x+h) - f(x)/h and f(x)-f(a)/x-a
I dont quit remember how these are done. I know that wherever x occurs you substitute in the -5x+2 but when i get (-5x +2)+h - (-5x+ 2)/h or -5x+2 - f(a)/-5x+2 I dont know where to go from here. I havent takes math like this for a few years and I'm struggling here. Please help.

I also have a piecewise function problem that I can't quite wrap my head around.
find the indicated values of f, the domain and range and the values of x in the domain of f at which f is discontinuous.

f(-2), f(1), f(2)
f(x) = x if -2 is lessthan or equal to x thats less than 1
-x +2 if 1 is lessthan or equal to x and is less than or equal to 2

........ for the values of this i got -2, 1, and 0 respectively but i am not sure how to graph it to find the domain and the range.
 
Princezz3286 said:
I havent takes math like this for a few years

You might need to back up and review.


\(\displaystyle f(x + h) = -5(x + h) + 2\)

\(\displaystyle f(x) = -5x + 2\)

Therefore:

\(\displaystyle \frac{f(x + h) - f(x)}{h} = \frac{-5(x + h) + 2 - (-5x + 2)}{h}\)

Now expand and simplify the numerator, and you will see a cancellation that leaves -5 as the answer.

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\(\displaystyle f(a) = -5a + 2\)

\(\displaystyle \frac{f(x) - f(a)}{x - a} = \frac{-5x + 2 - (-5a + 2)}{x - a}\)

Again, simplify the numerator. Factor out -5, and you will see a cancellation that again leaves -5 as the answer.

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To graph the piecewise function, plot the three points and connect them, using the given slopes as a guide.

[attachment=0:2o9ivrjk]piecewise.JPG[/attachment:2o9ivrjk]

 

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