Question

Let `f(x) = 5 + 3 x^2`. If h cannot equal 0, then the difference quotient can be simplified as (f(x+h)-f(x))/h = Ah + Bx + C,?

(1 pt) Let `f(x) = 5 + 3 x^2`. If h cannot equal 0, then the difference quotient can be simplified as
`(f(x+h)-f(x))/h = Ah + Bx + C`,
where A, B, and C are constants. (Note: It's possible for one or more of these constants to be 0.) Find the constants.
A =
, B =
, and C =
.

Use the simplified expression to find f'(x) = lim as h->0 (f(x+h)-f(x))/h =__

Finally, find each of the following:

f'(1) =
, f'(2) =
, and f'(3) =
.

Answer 1

Please see below.

Explanation:

Let's not let the problem overwhelm us. We'll just do one thing at a time and then do the next thing.

`f(x) = 5+3x^2` Now it says something about the difference quotient and how it can be simplified to the form Stop! Back up.

The difference quotient is

`(f(x+h)-f(x))/h` which we can write

`(f(x+h)-f(x))/h = (overbrace([5+3(x+h)^2])^(f(x+h)) - overbrace([5+3x^2])^(f(x)))/h`

Now let's simplify (a few steps):

`= ([5+3(x^2+2xh+h^2)]-[5+3x^2])/h`

`= (5+3x^2+6xh+3h^2-5-3x^2)/h`

`= (6xh+3h^2)/h`

Now, since we know that `h != 0`, we can reduce (after we factor)

`= (cancel(h)(6x+3h))/cancel(h)`

`= 6x+3h` and I don't see anything I can do to simplify more than that. So let's go back to the problem and read some more.

It says we can get the form `Ah+Bx+C` and asks us to say what `A`, `B`, and `C` are.

We can (obviously?) write our simplified expression as `3h+6x+0`, so we have

`A = 3`, `B = 6`, and `C = 0`

What's next?

Use the simplified expression (that is `6x+3h`) to find

`f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h`

Now we have already found that `(f(x+h)-f(x))/h = 6x+3h` for `h = 0`. And we know that the limit doesn't poay any attention to what actually happens at `0`, so

`f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h = lim_(hrarr0)(6x+3h)`
And if `h` is really, really clos to `0`, then `3h` is really close to `0` so `6x+3h` is really close to `6x`

`f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h = lim_(hrarr0)(6x+3h) = 6x`

`f'(x) = 6x`.

Good, what's next?

Ah, plug in `1`, `2`, and `3` to find `f'(1) = 6` and `f'(2) = 12` and `f'(3) = 18`.