Question

If the function f : R → R defined by f(x) = $\frac{3^{x} + 3^{-x}}{2}$, then show that f(x + y) + f(x - y) = 2 f(x) f(y)

Answer 1

Answer:

f(x+y) + f(x-y) = 2 f(x) f(y)

Step-by-step explanation:

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Answer 2

If the function f : R → R defined by
f(x) = $\frac{3^{x} + 3^{-x}}{2}$
than

$f(y) = \frac{ {3}^{y} + {3}^{ - y} }{2}$
So

$f(x + y) = \frac{ {3}^{x + y} + {3}^{ - x - y} }{2} \\ \\ f(x - y) = \frac{ {3}^{x - y} + {3}^{ - x + y} }{2} \\ \\ f(x + y) + f(x - y) = \frac{ {3}^{x + y} + {3}^{ - x - y} }{2} + \frac{ {3}^{x - y} + {3}^{ - x + y} }{2} \\ \\ = \frac{ {3}^{x} {3}^{y} +{3}^{ - x} {3}^{ - y} +{3}^{x} {3}^{ - y} + {3}^{ - x} {3}^{y} }{2} \\ \\ = \frac{ {3}^{x}( {3}^{y} + {3}^{ - y} ) + {3}^{ - x} ( {3}^{y} + {3}^{ - y}) }{2} \\ \\ = \frac{( {3}^{y} + {3}^{ - y})( {3}^{x} + {3}^{ - x})}{2} \\ \\ = 2\frac{( {3}^{y} + {3}^{ - y})}{2} \frac{( {3}^{x} + {3}^{ - x})}{2} \\ \\ = 2f(x)f(y)$
= R.H.S

hence proved