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(1)已知函数$f(x)=m\ln x+(m-1)x\ (m\in R )$,讨论\[f(x)\]的单调性;

(2)已知函数$f(x)=\frac{1}{3}a{{x}^{3}}-\frac{1}{2}(a+1){{x}^{2}}+x-2$,求函数$f(x)$的单调区间.

答案:

$m\le 0$时,$f\left( x \right)$$\left( 0\ \ \ +\infty \right)$单调递减;

当$0<m<1$时,$f\left( x \right)$在$\left( 0\ \ \ \frac{m}{1-m} \right)$单调递增,$f\left( x \right)$在$\left( \frac{m}{1-m}\ \ \ +\infty \right)$单调递减;

当$m\ge 1$时,$f\left( x \right)$在$\left( 0\ \ \ +\infty \right)$单调递增.

(2)

$a<0$时,$f(x)$的单调递增区间为$\left( \frac{1}{a}\ \ \ 1 \right)$,单调递减区间为$\left( -\infty \ \ \ \frac{1}{a} \right)$$(1\ \ \ +\infty )$

$a=0$时,$f(x)$的单调递增区间为$(-\infty \ \ \ 1)$,单调递减区间为$(1\ \ \ +\infty )$

$0<a<1$时,$f(x)$的单调递增区间为$\left( -\infty \ \ \ 1 \right)$$\left( \frac{1}{a}\ \ \ +\infty \right)$;单调递减区间为$\left( 1\ \ \ \frac{1}{a} \right)$

$a=1$时,$f(x)$的单调递增区间为$\left( -\infty \ \ \ +\infty \right)$

$a>1$时,$f(x)$的单调递增区间为$\left( -\infty \ \ \ \frac{1}{a} \right)$$\left( 1\ \ \ +\infty \right)$;单调递减区间为$\left( \frac{1}{a}\ \ \ 1 \right)$


解答:

$m\le 0$时,$f\left( x \right)$$\left( 0\ \ \ +\infty \right)$单调递减;

$0<m<1$时,$f\left( x \right)$$\left( 0\ \ \ \frac{m}{1-m} \right)$单调递增,$f\left( x \right)$$\left( \frac{m}{1-m}\ \ \ +\infty \right)$单调递减;

$m\ge 1$时,$f\left( x \right)$$\left( 0\ \ \ +\infty \right)$单调递增.

(2)


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