How do I find the exponential function of the form $f (x) = a b^{x}$ $f(x)=ab^x$ for which $f (- 1) = 10$ $f(-1)=10$ and $f (0) = 5$ $f(0)=5$ ?

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Answer 1 · 2015-01-30T21:33:38

The answer is: $y = 5 {(\frac{1}{2})}^{x}$ $y=5(1/2)^x$

$f (- 1) = 10 \Rightarrow 10 = a b^{- 1} \Rightarrow 10 = \frac{a}{b} \Rightarrow a = 10 b$ $f(-1)=10rArr10=ab^-1rArr10=a/brArra=10b$

$f (0) = 5 \Rightarrow 5 = a b^{0} \Rightarrow 5 = a \cdot 1 \Rightarrow a = 5$ $f(0)=5rArr5=ab^0rArr5=a*1rArra=5$ .

So:
$a = 5 and b = \frac{1}{2}$ $a=5andb=1/2$ .

The function is: $y = 5 {(\frac{1}{2})}^{x}$ $y=5(1/2)^x$ .

How do I find the exponential function of the form f(x)=abxf(x)=ab^x for which f(−1)=10f(-1)=10 and f(0)=5f(0)=5?