Show all work. PLEASE HELP!!!!

Mathematics

1 answer:

34kurt3 years ago

7 0

(a) Using the table, give the values fo rthe inverse

1) original table of values:

x      1     2     3    4    5

f(x)   0     1     1    5    3

2) The inverse of the function is obtained by exchanging x and f(x), this is:

                   ( x, f(x) ) → ( f(x), x)

3) So, the table of values of the inverse of the given function is:

x          0    1     1      5     3

f⁻¹ (x)   0     1     2     3     4

(b) Is the inverse a function?

No, the inverse is not a function, since the table of the inverse shows that the x -value 1 has two different images.

This ambigüity is opposite to the definition of a function, which requires that any input value has only one output. For that reason, the inverse is not a function. You cannot tell whether the image of 1 is 1 or 2, because both are images of the same value.

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In a previous exercise we formulated a model for learning in the form of the differential equation dp dt = k(m − p) where p(t) m

DaniilM [7]

I assume you mean

$\dfrac{dP}{dt} = k(M-P)$

ANSWER
An expression for P(t) is

$P = M - Me^{-kt}$

EXPLANATION
This is a separable differential equation. Treat M and k as constants. Then we can divide both sides by M - P to get the P term with the differential dP and multiply both sides by dt to separate dt from the P terms

$\begin{aligned} \dfrac{dP}{dt} &= k(M-P) \\ \dfrac{dP}{M-P} &= k\, dt
\end{aligned}$

Integrate both sides of the equation.

$\begin{aligned}
\int \dfrac{dP}{M-P} &= \int k\, dt \\
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\ln|M-P| &= -kt - C\end{aligned}$

Note that for the left-hand side, u-substitution gives us

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hence why $\int \frac{dP}{M-P} \ne \ln|M - P|$

Now we use the definition of the logarithm to convert into exponential form.

The definition is

$\ln(a) = b \iff \log_e(a) = b \iff e^b = a$

so applying it here, we get

$\begin{aligned} \ln|M-P| &= -kt - C \\ |M - P| &= e^{-kt - C} \\ 
M - P &= \pm e^{-kt - C} 
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Exponent properties can be used to address the constant C. We use $x^{a} \cdot x^{b} = x^{a+b}$ here:

$\begin{aligned}
 M - P &= \pm e^{-kt - C} \\
M - P &= \pm e^{- C - kt} \\ 
M - P &= \pm e^{- C + (- kt)} \\ 
M - P &= \pm e^{- C} \cdot e^{- kt} \\ 
M - P &= Ke^{- kt} && (\text{\footnotesize Let $K = \pm e^{-kt}$ }) \\ 
M &= Ke^{- kt} + P\\
P &= M - Ke^{- kt}
\end{aligned}$

If we assume that P(0) = 0, then set t = 0 and P = 0

$\begin{aligned} 
0 &= M - Ke^{- k\cdot 0} \\
0 &= M - K \cdot 1 \\
M &= K
 \end{aligned}
$

Substituting into our original equation, we get our final answer of

$P = M - Me^{-kt}$

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