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ss7ja [257]
3 years ago
6

Function f is an exponential function. x −1 1 3 5 7 f(x) 14 4 64 1024 16,234 By what factor does the output value increase as ea

ch input value increases by 1? Enter your answer in the box.
Mathematics
1 answer:
Maru [420]3 years ago
4 0

Function f is an exponential function.

x      −1     1     3          5           7

f(x)     14    4   64     1024     16,234

We use exponential function y =ab^x

We find out factor b

When x=1 then y=4, so equation becomes 4=ab^1

When x=3 then y=64, so equation becomes 64=ab^3

Divide the second equation by first equation

\frac{64=ab^3}{4=ab^1}

16 = b^2

Take square root on both sides

So b = +-4

f(x) is increasing, so b= 4

Answer is 4

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Write the polynomial in factored form as a product of linear factors f(r)=r^3-9r^2+17r-9
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Answer:

  f(r) = (x -1)(x -4+√7)(x -4-√7)

Step-by-step explanation:

The signs of the terms are + - + -. There are 3 changes in sign, so Descartes' rule of signs tells you there are 3 or 1 positive real roots.

The rational roots, if any, will be factors of 9, the constant term. The sum of coefficients is 1 -9 +17 -9 = 0, so you know that r=1 is one solution to f(r) = 0. That means (r -1) is a factor of the function.

Using polynomial long division, synthetic division (2nd attachment), or other means, you can find the remaining quadratic factor to be r^2 -8r +9. The roots of this can be found by various means, including completing the square:

  r^2 -8r +9 = (r^2 -8r +16) +9 -16 = (r -4)^2 -7

This is zero when ...

  (r -4)^2 = 7

  r -4 = ±√7

  r = 4±√7

Now, we know the zeros are {1, 4+√7, 4-√7), so we can write the linear factorization as ...

  f(r) = (r -1)(r -4 -√7)(r -4 +√7)

_____

<em>Comment on the graph</em>

I like to find the roots of higher-degree polynomials using a graphing calculator. The red curve is the cubic. Its only rational root is r=1. By dividing the function by the known factor, we have a quadratic. The graphing calculator shows its vertex, so we know immediately what the vertex form of the quadratic factor is. The linear factors are easily found from that, as we show above. (This is the "other means" we used to find the quadratic roots.)

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