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melamori03 [73]
3 years ago
13

What is the inverse of the function f(x) = one-quarterx – 12?

Mathematics
2 answers:
KATRIN_1 [288]3 years ago
8 0

Answer:

f^{-1} (x) = 4x + 48

Step-by-step explanation:

let y = f(x) and rearrange making x the subject, that is

y = \frac{1}{4} x - 12 ( multiply through bt 4 to clear the fraction )

4y = x - 48 ( add 48 to both sides )

4y + 48 = x

Change t back into terms of x with x = f^{-1} (x)

f^{-1} (x) = 4x + 48

LenaWriter [7]3 years ago
3 0

Answer:

y = 4x - 48

Step-by-step explanation:

y = 1/4 x - 12

x = 1/4 y - 12

4x = y - 48

y = 4x - 48

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Power Series Differential equation
KatRina [158]
The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for y

\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0

which indeed gives the recurrence you found,

a_{n+3}=-\dfrac{n-3}{n+3}a_n

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that a_2=0, and substituting this into the recurrence, you find that a_2=a_5=a_8=\cdots=a_{3k-1}=0 for all k\ge1.

Next, the linear term tells you that 6a_0+6a_3=0, or a_3=a_0.

Now, if a_0 is the first term in the sequence, then by the recurrence you have

a_3=a_0
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Finally, the quadratic term gives 6a_1-12a_4=0, or a_4=\dfrac12a_1. Then by the recurrence,

a_4=\dfrac12a_1
a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1
a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1
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and so on, such that

a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}

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Now, the solution was proposed to be

y=\displaystyle\sum_{n\ge0}a_nx^n

so the general solution would be

y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots
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y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)
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