If f (x)= cases 8^x-4^x-2^x+1 x^2 , & if x>0 e ^x x+x+ 4, & if x 0 cases is continuous at x=0 then the value of 1000 e ^ =

Step-by-step solution

Continuity of f(x) at x=0 requires that _ x 0^- f(x) = _ x 0^+ f(x) = f(0) . Evaluating f(0) from the left-hand expression: f(0) = e^0 0 + 0 + 4 = 4 . The left-hand limit matches this value: _ x 0^- (e^x x + x + 4) = 4 . For the right-hand limit, factor the numerator: 8^x - 4^x - 2^x + 1 = (4^x - 1)(2^x - 1) This gives _ x 0^+ f(x) = _ x 0^+ (4^x - 1)(2^x - 1) x^2 = ( 4^x - 1 x ) ( 2^x - 1 x ) Applying the standard limit _ x 0 a^x - 1 x = a yields _ x 0^+ f(x) = ( 4)( 2) Equating the limits for continuity: 4 = ( 4)