Question

Can someone please check my answers?

1. For the function ƒ(x) = 10x2 + x, evaluate and simplify the expression: f(a+h)-f(a) / h

I got f(a+h)-f(a) / h = 20a+1


2. Find ƒ−1 for the function f(x) = 23-3x/6x

I got f^-1 (x) = 23/6x-3

Answer 1

only f(a) is divided by h?

Answer 2

yes

Answer 3

how did you arrive at 20a+1?

Answer 4

to be honest i'm not really sure, I asked someone how to do it a while ago but I forgot how to do it

Answer 5

when I got the answer they told me I was right but I wanted to double check

Answer 6

I'm getting 10(a+h)^2+(a+h)-(10a^2+a)/h

Answer 7

can you tell me how your getting it so I can understand?

Answer 8

sure

Answer 9

Pretty sure it's \(\dfrac{f(a+h)-f(a)}{h}\)

This is sorta preparation for calculus lol.

Answer 10

Basically, the way i understand f(a+h) is 'replace all x in function f(x) with a+h.

So, that's the first thing I did.

Which gets me 10(a+h)^2+(a+h)

Then, I subtract f(a) (which is the same as f(x), instead we use a in place of x)

Which makes 10(a+h)^2+(a+h) - 10a^2+a

The last part needs to be divided by h, which makes:

\[10(a+h)^2+(a+h)- \frac{ 10a^2+a }{ h }\]

Which can't really be simplified any further.

Answer 11

@geerky42 that one makes more sense to me as well, that's why I asked.

Answer 12

ohh okay thank you zimmah

Answer 13

how about the 2nd question?

Answer 14

It doesn't look right, but I don't know what it is supposed to be

Answer 15

@geerky42 could you help him with the second question here, I'm stuck there.

Answer 16

are you sure you didn't make a typo there? It's really f(x) = 23-3x/6x ?

Answer 17

yes it is

Answer 18

http://www.wolframalpha.com/input/?i=inverse+function+23-3x%2F6x

Answer 19

Maybe this link can help you:

Scroll to the lightblue box

http://tutorial.math.lamar.edu/Classes/CalcI/InverseFunctions.aspx

Answer 20

okay thank you