If f (x) is continuous on its domain [-2,2] , where f (x)= cases a x x +3 & , for -2 x < 0 2 x+7 & , for 0 x 1 x^2+8 - b , & for 1 < x 2 cases then the value of 2 a+3 b is

Step-by-step solution

Since f (x) is continuous in [-2,2] , it is continuous at x=0 and x=1 . _ x 0⁻ f (x)= _ x 0⁺ f (x) _ x 0⁻ ( a x x +3 )= _ x 0⁺ (2 x+7) aligned & a+3=0+7 & a=4 aligned Also, _ x 1⁻ f (x)= _ x 1⁺ f (x) _ x 1⁻ (2 x+7)= _ x 1⁺ ( x^2+8 -b ) aligned & 2(1)+7= 1+8 -b & 9=3-b & b=-6 aligned 2 a+3 b=8-18=-10