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Let $f$ be a function from $\mathbb{R}$ to $\mathbb{R}$ and $m$ and $n$ real numbers such that $mf(x-1) + nf(-x) = 2|x|+1$ for all real numbers. If $f(-2) = 5$ and $f(1) = 1$ , then find the numbers $m$ and $n$

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Transformations of quadratic functions

Full solution

Q. Let

f

be a function from

\mathbb{R}

to

\mathbb{R}

and

m

and

n

real numbers such that

mf(x-1) + nf(-x) = 2|x|+1

for all real numbers. If

f(-2) = 5

and

f(1) = 1

, then find the numbers

m

and

n

Substitute $x = -2$ : Plug in $x = -2$ into the equation $mf(x-1) + nf(-x) = 2|x|+1$ . $\newline$ Calculation: $mf(-3) + nf(2) = 2|-2| + 1 = 5$ . $\newline$ Since $f(-2) = 5$ , substitute and simplify. $\newline$ $mf(-3) + 5n = 5$ .
Substitute $x = 1$ : Plug in $x = 1$ into the equation $mf(x-1) + nf(-x) = 2|x|+1$ . $\newline$ Calculation: $mf(0) + nf(-1) = 2|1| + 1 = 3$ . $\newline$ Since $f(1) = 1$ , substitute and simplify. $\newline$ $m + nf(-1) = 3$ .
Find $f(-3)$ and $f(-1)$ : We now have two equations: $\newline$ $1$ . $mf(-3) + 5n = 5$ $\newline$ $2$ . $m + nf(-1) = 3$ $\newline$ We need to find values for $f(-3)$ and $f(-1)$ to solve for $m$ and $n$ .
Assume $f$ is even: Assume $f$ is an even function since it satisfies the given functional equation involving $|x|$ . $\newline$ Calculation: $f(-x) = f(x)$ , so $f(-3) = f(3)$ and $f(-1) = f(1)$ . $\newline$ Since $f(1) = 1$ , $f(-1) = 1$ .
Substitute $f(-1) = 1$ : Substitute $f(-1) = 1$ into equation $2$ . $\newline$ Calculation: $m + n = 3$ .
Find $f(-3)$ or $f(3)$ : We need to find $f(-3)$ or $f(3)$ . Assume $f(3) = f(-3) = a$ . $\newline$ Substitute into equation $1$ : $ma + 5n = 5$ .
Solve equations simultaneously: We now have: $\newline$ $1$ . $ma + 5n = 5$ $\newline$ $2$ . $m + n = 3$ $\newline$ Solve these equations simultaneously.
Express $m$ in terms of $n$ : From equation $2$ , express $m$ in terms of $n$ : $m = 3 - n$ . $\newline$ Substitute into equation $1$ : $(3 - n)a + 5n = 5$ .

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