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Function fx) is defined asx-12x2-7x + 5fix) =, is f(x) differentiable at x = 1 if yes find f'(1)?x=13 |
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Answer» at x≠1 the limit of the function is. f(x) = (x-1)/2x²-7x+5factorisation of 2x²-7x+5 is (x-1)(2x-5)f(x) = (x-1)/(x-1)(2x-5) ... so, (x-1) will be cancelled.. and f(x) = 1/(2x-5).... so the value at x tends to 1 = 1/(2-5) = -1/3... and at x = 1 , it is as usual -1/3...so, the function is continuous..now differentiate f(x) = 1/(2x-5) .. we get f'(x).= -1.2/(2x-5)² , at x = 1 the value is -2/9... so, it is differentiable.. and value of f'(1) = -2/9 |
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