# Math Help - 3 question new

1. ## 3 question new

The other thread wasn't working so here are the problems again . Please help me and my brother this is due towmorrow for him.

1.
a) Let f(x) = 2x Find the first differences between the values of f for x= 0,1,2,3,4, and 5. Write a formula that gives the difference between f(x+1) and f(x) in terms of x.
b) Repeat part (a) but use function g(x)=3x
c) Repeat part (a) but use the function h(x) = 5x
d) Make a conjecture for a general formula that gives the difference between f(x+1) and f(x) for the function k(x) = ax , for a > 1.
2. The average growth rate of exponential function f(x) on an interval a≤x≤b is f(b) – f(a)
b-a
The instantaneous growth rate at a is the number that this fraction approaches (if any) as b → a. Let f(x)=3x, and let b=a+h. Write the instantaneous growth rate at a as the product of 3a and a function of h (that does not involve a). Use a calculator to estimate the number this second factor approaches as h → 0.

3. Let f(x) =bx . For a fixed positive integer n, let g(x)= [f(x)]n , and let h(x) = f(f(f(…(x)…)))

a) Express g(x) as f(?)
b) Express h(x) as f(?)

2. Let's see.
f(x) = 2x
Make a table:
Code:
  _x_      _f(x)_
0          0          f(0) = 0
1          2          f(1) = 2
2          4          f(2) = 4
3          6          f(3) = 6
4          8          f(4) = 8
5          10         f(5) = 10
So you can find the differences from here,
f(1) - f(0) = 2 - 0 = 2
f(2) - f(1) = 4 - 2 = 2
f(3) - f(2) = 6 - 4 = 2
...
As you see, all differences of f(x)s for consecutive x's are 2. But this is just a basic deduction, it's not proven yet.

a) In a function (for example, f(x) = 3x + 4) you can replace all x's with anything you want.

In your function ( f(x) = 2x ) we will replace all x's with x+1 to get a function for f(x+1).
f(x) = 2x
f(x+1) = 2(x+1)
f(x+1) = 2x + 2
Now we have functions f(x) = 2x and f(x+1) = 2x + 2

"Write a formula that gives the difference between f(x+1) and f(x) in terms of x"
So we are looking for f(x+1) - f(x)
f(x+1) - f(x) = (2x + 2) - (2x)
f(x+1) - f(x) = 2
Aha, all differences for consecutive x's are 2, we proved it now.

3. can someone help with questions 2 and 3

4. "The instantaneous growth rate at a is the number that this fraction approaches (if any) as b → a. Let f(x)=3x, and let b=a+h."

Is it $f(x) = 3x$ or $f(x) = 3^x$ ?
I'm asking because it says "The average growth rate of exponential function f(x)..."

5. it's
sorry

Ok i just looked at the problem and everywhere where it says 2x or 3x or bx it's supposed to be or whatever the number is

6. can someone help please

7. In 2. question, there's the formula for finding instantaneous growth rate. The limit of this formula is called derivative.

Instantaneous growth rate at the point a is,
$\lim_{h \to 0 } \frac {f(a+h) - f(a)}{h}$

You can't calculate this exactly without knowing how to find limits, but you can get approximate results by giving h approximate values.

Your function is $f(x) = 3^x$
$\frac {f(a+h) - f(a)}{h}$
$\frac {3^{a+h} - 3^a}{h}$
In here, h is a value near 0 (nearer, more accurate)

If you want to calculate the instantaneous growth rate at x = a = 1,
$\frac {3^{1+h} - 3^1}{h}$
Let's say h = 0.001
Then $\frac {3^{1+h} - 3^1}{h}$
$\frac {3^{1 + 0.001} - 3^1}{0.001}$
= 3,297..

Attached Graph:
$f(x) = 3^x$

9. 3. Let f(x) =bx . For a fixed positive integer n, let g(x)= [f(x)]n , and let h(x) = f(f(f(…(x)…)))

a) Express g(x) as f(?)
b) Express h(x) as f(?)

$f(x) = b^x$
$g(x) = n f(x)$
$h(x) = f(f(f(...x...)))$

(Is it g(x) = n.f(x) or [f(x)]^n? I made it for the first)
$g(x) = n f(x)$
$g(x) = n b^x$
$g(x) = b^{\log _bn}.b^x$
$g(x) = b^{\log _bn + x}$
$g(x) = f(\log _bn + x)$

$h(x) = f(f(f(...x...)))$
$h(x) = f(h(x))$

10. it's for the second thank you so much for your help but yay it's for the second