Or the function f given above, determine whether the following conditions are true. input t if the condition is ture, otherwise input f . (a) f′(x)< 0 if 00 if x> 2; f (c) f″(x)< 0 if 0≤x< 1; t (d) f″(x)> 0 if 14; t (f) two inflection points of f(x) are, the smaller one is x= 2 and the other is x= 2

Evaluate 8x2 + 9x − 1 2x3 + 3x2 − 2x dx. solution since the degree of the numerator is less than the degree of the denominator, we don't need to divide. we factor the denominator as 2x3 + 3x2 − 2x = x(2x2 + 3x − 2) = x(2x − 1)(x + 2). since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand has the form† 8x2 + 9x − 1 x(2x − 1)(x + 2) = correct: your answer is correct. to determine the values of a, b, and c, we multiply both sides of this equation by the product of the denominators, x(2x − 1)(x + 2), obtaining 8x2 + 9x − 1 = a correct: your answer is correct. (x + 2) + bx(x + 2) + cx(2x − 1).