Thursday, August 24, 2017

Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 30

a.) Suppose that f(t)=t2t, find f(t).

Using the definition of derivative




f(t)=limh0f(t+h)f(t)hf(t)=limh0(t+h)2t+h(t2t)hSubstitute f(t+h) and f(t)f(t)=limh0\cancelt2+2th+h2t+h\cancelt2+thExpand the equation and combine like termsf(t)=limh02th+h2t+h+thIsolate the terms that has square root and multiply it by its conjugatef(t)=limh02th+h2h+tt+hht+t+ht+t+hGet the factor of the first term and multiply both numerator and denominator of the second term by (t+t+h)f(t)=limh0\cancelh(2t+h)\cancelh+t\cancelt(t+h)+\cancelt(t+h)(t+h)h(t+t+h)Cancel out and combine like termsf(t)=limh02t+h+\cancelt\cancelthh(t+t+h)Combine like terms in the second termf(t)=limh02t+h\cancelh\cancelh(t+t+h)Cancel out like terms in the second termf(t)=limh0(2t+h1t+t+h)=2t+01t+t+0=2t1t+tEvaluate the limitf(t)=2t12t


b.) Compare the graphs of f and f and check whether your answer in part (a) is reasonable.

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