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Let f(x)=2^(x) and g(x)=x^(2)+4. What is the value of f(1+g(0)) ?

Let f(x)=2x f(x)=2^{x} and g(x)=x2+4 g(x)=x^{2}+4 . What is the value of f(1+g(0)) f(1+g(0)) ?

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Full solution

Q. Let f(x)=2x f(x)=2^{x} and g(x)=x2+4 g(x)=x^{2}+4 . What is the value of f(1+g(0)) f(1+g(0)) ?
  1. Calculate g(0)g(0): We are given two functions:\newlinef(x)=2xf(x) = 2^{x}\newlineg(x)=x2+4g(x) = x^{2} + 4\newlineWe need to find the value of f(1+g(0))f(1+g(0)).\newlineFirst, we will calculate g(0)g(0).
  2. Calculate 1+g(0)1 + g(0): Substitute x=0x = 0 into g(x)g(x) to find g(0)g(0).
    g(0)=02+4g(0) = 0^{2} + 4
    g(0)=0+4g(0) = 0 + 4
    g(0)=4g(0) = 4
    Now we have the value of g(0)g(0).
  3. Substitute into f(x)f(x): Next, we will calculate 1+g(0)1 + g(0).
    1+g(0)=1+41 + g(0) = 1 + 4
    1+g(0)=51 + g(0) = 5
    Now we have the value of 1+g(0)1 + g(0).
  4. Calculate 252^{5}: Now we will substitute 1+g(0)1 + g(0) into f(x)f(x) to find f(1+g(0))f(1+g(0)).\newlinef(1+g(0))=f(5)f(1+g(0)) = f(5)\newlinef(5)=25f(5) = 2^{5}
  5. Calculate 252^{5}: Now we will substitute 1+g(0)1 + g(0) into f(x)f(x) to find f(1+g(0))f(1+g(0)).\newlinef(1+g(0))=f(5)f(1+g(0)) = f(5)\newlinef(5)=25f(5) = 2^{5}Calculate 252^{5}.\newline25=322^{5} = 32\newlineSo, f(1+g(0))=32f(1+g(0)) = 32.

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