Question 187097
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The first thing to realize is that if <i>f</i>( <i>x</i> ) were non-linear, then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x+2) – f(x-3)]


would be some function of <i>x</i> rather than a constant.  Hence <i>f</i>( <i>x</i> ) is linear.


Given that, we can use the point-slope form of the equation of a line to derive an equation for the line.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y - 6 = m(x + 2) ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x) = y = mx + 2m + 6]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x+2) =  m(x+2) + 2m + 6 = mx + 4m + 6 ]


And:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x-3) =  m(x-3) + 2m + 6 = mx - m + 6]


And then:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x+2) – f(x-3) = (mx + 4m + 6) - (mx - m + 6) = 5m = 20]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m = 4]


Now:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x) = 4x + 8 + 6 = 4x + 14]


Check the answer:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(-2) = 4(-2) + 14 = 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x + 2) = 4(x + 2) + 14 = 4x + 22]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x - 3) = 4(x - 3) + 14 = 4x + 2]


and 22 - 2 = 20.
 

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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