Can someone help me with these problems.

[j9vo2]

(1.) given f(x) = 4 + 3x - x ^ 2, find and simplify f(a+h) - f(a) I have no idea how to even begin this one.


(2.) this one is absolute value: |8-q| = |q +19|.. I tried to figure this one out and I got -11/2.. is this correct? the other two options were.. no solution, or -11/2, 0


(3.) 16 <= (less than or equal to) |2x-3| + 9. The answer I got for this one was
X <+ -2 AND x >= 5.. or should it be X <= -2 OR x > = 5

Thanks so much

 

[tkhunny]

(1.) given f(x) = 4 + 3x - x ^ 2, find and simplify f(a+h) - f(a) I have no idea how to even begin this one.

Sorry, but I simply dont believe that. What does the notation, f(x) mean? Substitute and go!!!

f(a) = 4 + 3a - a ^ 2
f(a+h) = 4 + 3(a+h) - (a+h) ^ 2

Now what?

 

[Mrspi]

(
(2.) this one is absolute value: |8-q| = |q +19|.. I tried to figure this one out and I got -11/2.. is this correct? the other two options were.. no solution, or -11/2, 0


(3.) 16 <= (less than or equal to) |2x-3| + 9. The answer I got for this one was
X <+ -2 AND x >= 5.. or should it be X <= -2 OR x > = 5

Thanks so much

For #2:

If |a| = |b|, then a^2 = b^2
Since |8 - q| = |q + 19|, it must be true that
(8 - q)<SUP>2</SUP> = (q + 19)<SUP>2</SUP>

Square each of the binomials and solve for q. Question....did you check the answer you got?? You should be able to determine for yourself whether your answer is correct.

For #3: hmmmmm....I don't think it is possible for both x <= -2 AND x >= 5 to be true.......

16 <= | 2x - 3 | + 9
7 <= |2x - 3|

So, 2x - 3 >= 7, OR 2x - 3 <= -7

Can you see what the correct answer should be now?

I hope this helps you.

 

[Gene]

1) For functions you replace the x in
f(x) = 4 + 3x - x ^ 2
with whatever is in the f(?).
f(?) = 4+3(?)-(?^2)
f(a+h) = 4 +3(a+h)-(a+h)^2
f(a) = 4 +3a-a^2

2) correct!

Brunch is ready. I'll be back.

 

[j9vo2]

Ok.. what does ^ mean.. I never saw this symbol before in my book. I am still trying to figure this one out. When you did f(a) = 4 + 3a-a^2.. would I do the same for f(h) (i.e) 4 + 3h -h^2.. ???


number 2 is correct... thanks.. I did substitute and I got 27/2 = 27/2

number 3.. I guess should be or than b/c -2 does not work in the problem. is this correct now?

 

[Gene]

^ is a standard symbol for exponent.
x^3 = x*x*x

#3) -2 does work
so does 5
so do -3 and 6
0 doesnt so it is
x<-2 or x>5, your second choice.

Yes, f(h) would be 4+3h-h^2 but
f(x+h) would be what you were told a couple times.
--------------------
Gene

 

[TchrQbic]

Ok.. what does ^ mean.. I never saw this symbol before in my book.

^ is used to indicate an exponent when using a keyboard. Or LaTex can be used to enter exponents in this forum. Thus, x^2 means x².

 

[j9vo2]

ok.. I think I am halfway there with this problem.. I got the following answer..

6a+3h-2a^2-h^2

now what do i do?

 

[Gene]

You aren't showing your work but
f(a+h) - f(a) =
[4+3(a+h)-(a+h)^2]-[4+3a-a^2] =
4+3a+3h-a^2-2ah-h^2-4-3a+a^2 =
3h-2ah-h^2 =
h(3-2a-h)

That's all you can do with it. Usually it would be
(f(a+h) - f(a))/h
then take the limit as h=>0 for the tangent at a

 

[j9vo2]

thanks soo much!!!