1.

Iff: R → R,f(x) = x2-5x + 7, then find the value off"(1).

Answer»

f-¹(1) , means the value of x , for which the function will have the value of 1.

=> x²-5x+7 = 1 => x²-5x+7-1 = 0 => x²-5x+6 = 0 => x²-3x-2x+6 = 0 => x(x-3) -2(x-3) = 0 => (x-3)(x-2) = 0 ==> x = 3,2.



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