Write a recrusive rule for the exponential functionf(x)=0.25(2/3)^x
1 answer:
Answer:
f(0) = 0.25 = 1/4
f(x) = f(x-1) × 2/3
Step-by-step explanation:
a0 = 0.25×(2/3)⁰ = 0.25×1 = 0.25 = 1/4
a1 = 0.25×(2/3)¹ = 0.25×(2/3) = 1/4 × 2/3 = 2/12 = 1/6 =
= a0 × 2/3
a2 = 0.25×(2/3)² = 0.25×(4/9) = 1/4 × 4/9 = 4/36 = 1/9 =
= a1 × 2/3
=>
an = an-1 × 2/3, n >= 1
but what about negative x ?
is a0 = a-1 × 2/3 ?
so, yes, a0 = a-1 × (2/3)
and a-1 = a-2 × (2/3)
...
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The answers are shown in the attached image
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Explanation:
Set the denominator x^4-8x^3+16x^2 equal to zero and solve for x
x^4-8x^3+16x^2 = 0
x^2(x^2-8x+16) = 0
x^2(x-4)^2 = 0
x^2 = 0 or (x-4)^2 = 0
x = 0 or x-4 = 0
x = 0 or x = 4
The x values 0 and 4 make the denominator zero
These x values lead to asymptote discontinuities because the numerator 8x-24 = 8(x-3) has no common factors which cancel with the denominator factors.
There are two vertical asymptotes
Let's see what happens when we plug in a value to the left of x = 0, say x = -1, we'd get
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(-1) = (8(-1)-24)/((-1)^4-8(-1)^3+16(-1)^2)
f(-1) = -1.28
So as x gets closer and closer to x = 0 from the left side, the f(x) is heading to negative infinity
Now plug in some value to the right of x = 0. I'm going to pick x = 1
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(1) = (8(1)-24)/((1)^4-8(1)^3+16(1)^2)
f(1) = -1.78 (approximate)
So as x gets closer and closer to x = 0 from the right side, the f(x) is heading to negative infinity
Overall, as x approaches 0 from either the left or right side of x = 0, the y value is heading off to negative infinity
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Repeat for values to the left and right of x = 4
We can't use x = 1 as it turns out that x = 3 is a root
But we can use something like x = 3.5 to find that...
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(3.5) = (8(3.5)-24)/((3.5)^4-8(3.5)^3+16(3.5)^2)
f(3.5) = 1.31 approx
So as x gets closer to x = 4 from the left, y is getting closer to positive infinity
Plug in x = 5 to find that
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(5) = (8(5)-24)/((5)^4-8(5)^3+16(5)^2)
f(5) = 0.64
which has the same behavior as the left side
So overall, as we approach x = 4, the y value is heading off to positive infinity
Again everything is summarized in the image attachment
Note: you could make a table of more values but they would effectively say what has already been said. It would be redundant busy work. However, its always good practice for function evaluation.
-------------------------------------------------------------------------
Explanation:
Set the denominator x^4-8x^3+16x^2 equal to zero and solve for x
x^4-8x^3+16x^2 = 0
x^2(x^2-8x+16) = 0
x^2(x-4)^2 = 0
x^2 = 0 or (x-4)^2 = 0
x = 0 or x-4 = 0
x = 0 or x = 4
The x values 0 and 4 make the denominator zero
These x values lead to asymptote discontinuities because the numerator 8x-24 = 8(x-3) has no common factors which cancel with the denominator factors.
There are two vertical asymptotes
Let's see what happens when we plug in a value to the left of x = 0, say x = -1, we'd get
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(-1) = (8(-1)-24)/((-1)^4-8(-1)^3+16(-1)^2)
f(-1) = -1.28
So as x gets closer and closer to x = 0 from the left side, the f(x) is heading to negative infinity
Now plug in some value to the right of x = 0. I'm going to pick x = 1
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(1) = (8(1)-24)/((1)^4-8(1)^3+16(1)^2)
f(1) = -1.78 (approximate)
So as x gets closer and closer to x = 0 from the right side, the f(x) is heading to negative infinity
Overall, as x approaches 0 from either the left or right side of x = 0, the y value is heading off to negative infinity
---------------------
Repeat for values to the left and right of x = 4
We can't use x = 1 as it turns out that x = 3 is a root
But we can use something like x = 3.5 to find that...
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(3.5) = (8(3.5)-24)/((3.5)^4-8(3.5)^3+16(3.5)^2)
f(3.5) = 1.31 approx
So as x gets closer to x = 4 from the left, y is getting closer to positive infinity
Plug in x = 5 to find that
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(5) = (8(5)-24)/((5)^4-8(5)^3+16(5)^2)
f(5) = 0.64
which has the same behavior as the left side
So overall, as we approach x = 4, the y value is heading off to positive infinity
Again everything is summarized in the image attachment
Note: you could make a table of more values but they would effectively say what has already been said. It would be redundant busy work. However, its always good practice for function evaluation.