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Let f be a function. Consider the function A(t) = : f (x) dx, which gives the average of f on the Exercise. interval (0, t Suppose that A has a local maximum at t = b, then A'(b) = the formula for A'(t) is 0 (answer with a number). Now A' (t) 2 Now since A, (b) = 0 A(b) = Now suppose that f(t) = A(t) for all t > 0, then A,(t) = (answer with a number), we can conclude that (answer with an expression in terms of f) 0 for all t> 0 and so A must be constant, that is, there exists c such that A(t) = f(x)dz = c So f(x)dx = to Differentiating both sides with respect to t we find that f(t)-f(0) that is, f is constant. Suppose f(x) = x(1-2). For what value of t is A(t) maximized? 2