Lim y t y t y y t 1 t lim y t g y g t y t lim y t g y

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= lim y t + ( y - t ) /y y - t = 1 t lim y t - g ( y ) - g ( t ) y - t = lim y t - g ( y ) y - t = lim y t - ( t - y ) /y y - t = - 1 t Since these are not equal to each other, the function g ( x ) is not differentiable at x = t . So x = t is a critical point, and by the first derivative test for local extrema, x = t is a local minima. Since there is no other point which is a critical point, this means that the maxima must occur at the end-points of the interval. (b) Part (a) tells us that M ( t ) = max t - a a , b - t b Plot the graphs of f 1 and f 2 , where f i are straight lines defined by f 1 ( t ) = t - a a , f 2 ( t ) = b - t b Note that these lines intersect at the point ( τ, f 1 ( τ )) where τ - a a = b - τ b If you do some algebra you find that at the point of intersection 1 τ = 1 2 1 a + 1 b Note that now we can say that M ( t ) = ( b - t b if t [ a, τ ] t - a a if t [ τ, b ] Hence (for example by the first derivative test, or by just sketching the graphs of f 1 and f 2 ), we see that the minimum of M ( t ) occurs at t = τ . 3

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6. Let f ( x ) = x 2 + x sin x + cos x . We will show that there are only two points where f ( x ) = 0. Note that when | x | is large, f ( x ) > 0 and f (0) = - 1. Since f is continous everywhere on the real line, this means that there are at least two points where f ( x ) = 0. We now show that there cannot be more than two points. Let us compute Df ( x ) = f 0 ( x ) = 2 x + x cos x Note that the derivative never goes to zero except at x = 0. If there were more than
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