# Given f(x)=max(x+2,2^x), what are f(0),f(1),f(2), f(k), if k greater 3 and what k=? if f(2k)-7f(k)-8=0?

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The function f(x)=max(x+2,2^x)

f(0) = 2, f(1) = 3, f(2) = 4. If k > 3, f(k) = 2^k

The value of k that satisfies f(2*k) - 7*f(k) - 8 = 0 has to be determined.

As the function returns different values for different values of k take the case where k lies in [-1, 2]

f(2*k) - 7*f(k) - 8 = 0

=> 2k + 2 - 7*(k + 2) - 8 = 0

=> 2k + 2 - 7k - 14 - 8 = 0

=> -5k = 20

=> k = -4

This does not lie in the required range.

If k > 2

f(2*k) - 7*f(k) - 8 = 0

=> 2^(2k) - 7*2^k - 8 = 0

let 2^k = x

=> x^2 - 7x - 8 = 0

=> x^2 - 8x + x - 8 = 0

=> x(x - 8) - 1(x - 8) = 0

=> k = 0 and k = 3

Of these only k = 3 is applicable

**The value of k for which f(2*k) - 7*f(k) - 8 = 0 is k = 3**