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受欢迎的函数 >

extreme C(t)= t/(4t^2+11)

解答

极值点

C (t) = \frac{t}{4 t ^{2} + 11}

解答

极小值 (- \frac{11}{2}, - \frac{1}{4 11}), 极大值 (\frac{11}{2}, \frac{1}{4 11})

求解步骤

f^{'} (t) = \frac{- 4 t ^{2} + 11}{( 4 t ^{2} + 11 ) ^{2}}

Find intervals:

递减

: - \infty < t < - \frac{11}{2},

递增

: - \frac{11}{2} < t < \frac{11}{2},

递减

: \frac{11}{2} < t < \infty

将

t = - \frac{11}{2}

代入

\frac{t}{4 t ^{2} + 11} : - \frac{1}{4 11}

极小值 (- \frac{11}{2}, - \frac{1}{4 11})

将

t = \frac{11}{2}

代入

\frac{t}{4 t ^{2} + 11} : \frac{1}{4 11}

极大值 (\frac{11}{2}, \frac{1}{4 11})

极小值 (- \frac{11}{2}, - \frac{1}{4 11}), 极大值 (\frac{11}{2}, \frac{1}{4 11})

作图

流行的例子

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