x A family of super-exponential functions? Let f(x) = (a? +x) ?, wher ? e ? > 0. a.? What is the dom ? ain of f ? (in te?rms of ?a)? b.? Describe the end be ? havior of f ? (near the boundary of its doma ? in or a? s ??x? ? ? ? ). ? c.? Compute ?f?.? Then? graph ? ?f and f? ? for ?a = 0.5, 1, 2, 3. ? d.? Show that ?f has a single local minimu?m at the point ?z that satisfies ? ? ? ? (?? + ?a)ln (?z + ?a)+ ?z = 0. ? e.? Describe how z ? [found in? part (d)] varies as ?a increases. Describe how f(z)varies as ?a increases.

Solution 76RE Step 1 In this problem we are given that f(x) = (a+x) where a > 0. We have to find the domain of f in terms of a and we have to explain the end behavior of f near the boundary. Then we need to find f and we need to draw the graph of f and f . Then we have to show that f has only one local minimum at the point z that satisfies the given condition. Step 2 a. Wha t is the domain o (in terms of ) Consider f(x) = (a+x) x Take the natural logarithm(ln) on both sides, we get ln f(x) = x ln (a+x) We know that limln x = x Thus whatever the higher limit the function exists. Hence its upper limit is We know that ln 0 does not exists. Thus the logarithmic function does not exists if the function value is 0. That is it does not exists ifxcm ln (xc) Thus, the lower limit of f(x) is a+x = 0 x = a Hence the domain of f (in terms of a ) is (a,) Step 3 b. Describe the end behavior of f (near the boundary of its domain or as x Here we have to explain the behaviour of f at the endpoints. That is at x = aand x = First let us consider the limit of f(x)as x a. x x af(x) = x a(a+x) x = lim (a+x) x+a0 = 0 x = 0 Now let us consider the limit of f(x)as x . limf(x) = lim (a+x) x x x = Hence the end behavior of f(x)is lim f(x) = 0and limf(x) = x a x Step 4 c. Compute f. Then graph f and f for a = 0.5, 1, 2, 3. x We have f(x) = (a+x) First let us find f (x) x f(x) = (a+x) Taking natural logarithm(ln) on both sides we get x ln f(x) = ln (a+x) ln f(x) = x ln (a+x) Now differentiating both sides with respect to xwe get, d d dx ln f(x) = dx(x ln (a+x)) 1 . f(x) = ln (a+x)+x. 1 f(x) a+x f(x= ln (a+x)+ x f(x) a+x x f (x) = f(x)[ln (a+x)+ a+x] We have f(x) = (a+x) . Substituting this we get, f(x) = (a+x) [ln (a+x)+ x ] a+x x x x Thus we have f(x) = (a+x) and f (x) = (a+x) [ln (a+x)+ a+x ]