How do you find formulas for the exponential functions satisfying the given conditions f(3)=-3/8 and f(-2)=-12?

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Answer 1 · 2016-10-19T23:08:56

$f (x) = - 3 {(\frac{1}{2})}^{x}$ $f(x) = -3(1/2)^x$

Suppose we have an exponential function of the form $f (x) = A e^{B x}$ $f(x) = Ae^(Bx)$

Given our conditions, we get the system

${(Ae^(3B) = -3/8), (Ae^(-2B)=-12):}$

Multiplying the first equation by $e^(-3B)$ and the second by $e^(2B)$ , we get

${(A = -3/8e^(-3B)), (A = -12e^(2B)):}$

Thus, equating the two, we have

$-3/8e^(-3B) = -12e^(2B)$

Multiplying by $e^(3B)/-12$ , we get

$1/32 = e^(5B)$

Taking the natural log of both sides, we find

$ln(1/32) = ln(e^(5B))$

$=> ln(2^(-5)) = 5B$

$=> -5ln(2) = 5B$

$=> B = -ln(2) = ln(1/2)$

Note, then, that $e^(Bx) = (e^B)^x = (e^(ln(1/2)))^x = (1/2)^x$

Now we can substitute this back into a prior equation, say, $A=-12e^(2B)$ . Doing so, we get

$A = -12(1/2)^2$

$=-12/4$

$=-3$

Thus we get our full equation as

$f(x) = Ae^(Bx) = -3(1/2)^x$

Checking we find that

$f(3) = -3(1/2)^3 = -3(1/8) = -3/8$
and
$f(-2) = -3(1/2)^(-2) = -3(4) = -12$

as desired.