Question

functions
f(x)=x^2-x+2,find a) f(0) , b).f(-1), c) f(x+h), and d). (f(x+h)-f(x))/h

Answer 1

f(0) means replace x with 0 in the equation, and then do the arithmetic. Try it.

Answer 2

phi is that the only thing am to do there?

Answer 3

that is (a)

Answer 4

ok

Answer 5

and the same thing in b?

Answer 6

what did you get for (a) and (b)?

Answer 7

i got 2 for both questions

Answer 8

Now (c), replace x with (x+h) in the equation, and expand things:
(x+h)^2 - (x-h) +2 = x^2 +2xh+h^2 -x +h +2

Answer 9

Now (d). f(x+h) is as above. f(x) is the original equation. Subtract them:

x^2 +2xh+h^2 -x +h +2 - (x^2-x+2)= ??
and then divide your answer by h.

Answer 10

am hooked,can u help me expand f(x+h)

Answer 11

?

Answer 12

first, you know f(x+h)= (x+h)^2 - (x-h) +2 (all x's replace with (x+h) )

I already expanded it: (x+h)^2 = x^2 +2xh+h^2
and -(x-h) is -x +h
so it becomes
x^2 +2xh+h^2-x +h+2

Unless you are asking how to expand (x+h)^2 ?

Answer 13

ok don't worry i was thinking we had to simply further from there

Answer 14

so that's my answer for c?

Answer 15

yes

Answer 16

cool thanks alot

Answer 17

hey what is the value of my h in dis case,when i want to divide

Answer 18

For (d) you have:
f(x+h) - f(x) = x^2 +2xh+h^2 -x +h +2 - (x^2-x+2)
this simplifies to
2xh +h +h^2

now, as long as h is not zero, we can divide to get
\( \frac{2xh+h+h^{2}}{h}= 2x+1 +h \)

In calculus, we let h approach zero (very,very close), so the expression approaches 2x+1

Answer 19

\[\frac{2xh+h+h^2}{h}= 2x+1+h\]