find a formula for the function

Question

e^-((x-a)^2/b) when b > 0 and with a local maximum at x=2 and points of inflection at x=1 and x=3

Answer 1

If you were a probability or statistics student, you would recognize this function as being in the form of the normal distribution. Thus, we see a mean of 2 and a standard deviation of 1 (to get inflections at 1 and 3)

the exponent for the normal is of the form

(x - mean)^2/2*standard deviation^2

Thus, as we have (x-a)^2/b, a = mean, or a = 2; b = 2 * standard deviation^2 or 2 * 1^2 = 2

If a calculus student, you take the first derivative and get

f'(x) = -2(x-a)/be^-(x-a)^2/b

x=a at the max; the max is 2, so a = 2

f''(x) = (-2/b+4(x-a)^2/b^2)e^-(x-a)^2/b

f''(x) = 0, so -2/b+4(x-a)^2/b^2 = 0

2/b = 4(x-a)^2/b^2

cross multiplying 4b (x-a)^2 = 2b^2

b = 2(x-a)^2

As a = 2, and x = 3 or 1, |x-a|= 1

Thus, b = 2*1^2

b = 2

We get the same answer from calculus that we get from a knowledge of statistics.