高中數(shù)學(xué)函數(shù)易錯(cuò)題型全解試卷（2025年）一、選擇題（本大題共10小題，每小題5分，共50分）1.已知函數(shù)$f(x)=2x^3-3x^2+4x-1$，則下列結(jié)論正確的是：（1）$f(x)$在$R$上單調(diào)遞增（2）$f(x)$在$R$上單調(diào)遞減（3）$f(x)$在$R$上有極值點(diǎn)（4）$f(x)$在$R$上無極值點(diǎn)2.已知函數(shù)$f(x)=\frac{x}{x+1}$，則下列結(jié)論正確的是：（1）$f(x)$在$(-\infty,0)$上單調(diào)遞增（2）$f(x)$在$(0,+\infty)$上單調(diào)遞減（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增（4）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞減3.已知函數(shù)$f(x)=\log_2(x+1)$，則下列結(jié)論正確的是：（1）$f(x)$在$(-1,+\infty)$上單調(diào)遞增（2）$f(x)$在$(-\infty,-1)$上單調(diào)遞減（3）$f(x)$在$(-\infty,-1)\cup(-1,+\infty)$上單調(diào)遞增（4）$f(x)$在$(-\infty,-1)\cup(-1,+\infty)$上單調(diào)遞減4.已知函數(shù)$f(x)=\frac{1}{x^2+1}$，則下列結(jié)論正確的是：（1）$f(x)$在$(-\infty,0)$上單調(diào)遞增（2）$f(x)$在$(0,+\infty)$上單調(diào)遞減（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增（4）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞減5.已知函數(shù)$f(x)=\sqrt{x^2+1}$，則下列結(jié)論正確的是：（1）$f(x)$在$(-\infty,0)$上單調(diào)遞增（2）$f(x)$在$(0,+\infty)$上單調(diào)遞減（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增（4）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞減6.已知函數(shù)$f(x)=x^3-x$，則下列結(jié)論正確的是：（1）$f(x)$在$(-\infty,0)$上單調(diào)遞增（2）$f(x)$在$(0,+\infty)$上單調(diào)遞減（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增（4）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞減7.已知函數(shù)$f(x)=\frac{x}{x^2+1}$，則下列結(jié)論正確的是：（1）$f(x)$在$(-\infty,0)$上單調(diào)遞增（2）$f(x)$在$(0,+\infty)$上單調(diào)遞減（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增（4）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞減8.已知函數(shù)$f(x)=\log_2(x+1)$，則下列結(jié)論正確的是：（1）$f(x)$在$(-\infty,0)$上單調(diào)遞增（2）$f(x)$在$(0,+\infty)$上單調(diào)遞減（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增（4）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞減9.已知函數(shù)$f(x)=\sqrt{x^2+1}$，則下列結(jié)論正確的是：（1）$f(x)$在$(-\infty,0)$上單調(diào)遞增（2）$f(x)$在$(0,+\infty)$上單調(diào)遞減（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增（4）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞減10.已知函數(shù)$f(x)=x^3-x$，則下列結(jié)論正確的是：（1）$f(x)$在$(-\infty,0)$上單調(diào)遞增（2）$f(x)$在$(0,+\infty)$上單調(diào)遞減（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增（4）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞減二、填空題（本大題共5小題，每小題5分，共25分）11.已知函數(shù)$f(x)=2x^3-3x^2+4x-1$，則$f'(x)=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\四、解答題（本大題共2小題，每小題15分，共30分）12.已知函數(shù)$f(x)=x^3-3x^2+4x-1$，求$f(x)$的極值點(diǎn)，并判斷極值的類型。要求：（1）求出$f(x)$的導(dǎo)數(shù)$f'(x)$；（2）令$f'(x)=0$，解得$x$的值；（3）判斷$x$的值在$f(x)$的定義域內(nèi)的單調(diào)性；（4）根據(jù)單調(diào)性判斷$f(x)$的極值點(diǎn)，并求出極值。13.已知函數(shù)$f(x)=\frac{1}{x^2+1}$，求$f(x)$的導(dǎo)數(shù)$f'(x)$，并分析$f(x)$在$(-\infty,+\infty)$上的單調(diào)性。要求：（1）利用導(dǎo)數(shù)的定義求出$f(x)$的導(dǎo)數(shù)$f'(x)$；（2）分析$f'(x)$的符號(hào)，判斷$f(x)$的單調(diào)遞增區(qū)間和單調(diào)遞減區(qū)間；（3）給出$f(x)$的單調(diào)性結(jié)論，并解釋理由。五、證明題（本大題共1小題，共15分）14.證明：若函數(shù)$f(x)=x^3-3x^2+4x-1$在$x=1$處取得極大值，則$f(x)$在$x=1$處的導(dǎo)數(shù)為0。要求：（1）求出$f(x)$的導(dǎo)數(shù)$f'(x)$；（2）利用導(dǎo)數(shù)的定義證明$f'(1)=0$；（3）結(jié)合$f(x)$的單調(diào)性和極值點(diǎn)的關(guān)系，證明結(jié)論。六、應(yīng)用題（本大題共1小題，共15分）15.已知函數(shù)$f(x)=\frac{1}{x+1}$，求證：$f(x)$在$(-\infty,-1)\cup(-1,+\infty)$上無極值點(diǎn)。要求：（1）求出$f(x)$的導(dǎo)數(shù)$f'(x)$；（2）分析$f'(x)$的符號(hào)，判斷$f(x)$的單調(diào)遞增區(qū)間和單調(diào)遞減區(qū)間；（3）結(jié)合$f(x)$的單調(diào)性和極值點(diǎn)的定義，證明結(jié)論。本次試卷答案如下：一、選擇題1.（3）$f(x)$在$R$上有極值點(diǎn)解析：函數(shù)$f(x)=2x^3-3x^2+4x-1$的導(dǎo)數(shù)為$f'(x)=6x^2-6x+4$，令$f'(x)=0$，解得$x=\frac{1}{2}$，$x=1$。當(dāng)$x<\frac{1}{2}$或$x>1$時(shí)，$f'(x)>0$，函數(shù)單調(diào)遞增；當(dāng)$\frac{1}{2}<x<1$時(shí)，$f'(x)<0$，函數(shù)單調(diào)遞減。因此，$x=\frac{1}{2}$和$x=1$是$f(x)$的極值點(diǎn)。2.（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增解析：函數(shù)$f(x)=\frac{x}{x+1}$的導(dǎo)數(shù)為$f'(x)=\frac{1}{(x+1)^2}$，由于$(x+1)^2>0$，所以$f'(x)>0$，函數(shù)在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增。3.（1）$f(x)$在$(-1,+\infty)$上單調(diào)遞增解析：函數(shù)$f(x)=\log_2(x+1)$的導(dǎo)數(shù)為$f'(x)=\frac{1}{(x+1)\ln2}$，由于$x+1>0$，$\ln2>0$，所以$f'(x)>0$，函數(shù)在$(-1,+\infty)$上單調(diào)遞增。4.（2）$f(x)$在$(0,+\infty)$上單調(diào)遞減解析：函數(shù)$f(x)=\frac{1}{x^2+1}$的導(dǎo)數(shù)為$f'(x)=-\frac{2x}{(x^2+1)^2}$，當(dāng)$x>0$時(shí)，$f'(x)<0$，函數(shù)在$(0,+\infty)$上單調(diào)遞減。5.（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增解析：函數(shù)$f(x)=\sqrt{x^2+1}$的導(dǎo)數(shù)為$f'(x)=\frac{x}{\sqrt{x^2+1}}$，當(dāng)$x>0$或$x<0$時(shí)，$f'(x)>0$，函數(shù)在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增。6.（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增解析：函數(shù)$f(x)=x^3-x$的導(dǎo)數(shù)為$f'(x)=3x^2-1$，令$f'(x)=0$，解得$x=\pm\frac{\sqrt{3}}{3}$。當(dāng)$x<-\frac{\sqrt{3}}{3}$或$x>\frac{\sqrt{3}}{3}$時(shí)，$f'(x)>0$，函數(shù)單調(diào)遞增；當(dāng)$-\frac{\sqrt{3}}{3}<x<\frac{\sqrt{3}}{3}$時(shí)，$f'(x)<0$，函數(shù)單調(diào)遞減。因此，$x=-\frac{\sqrt{3}}{3}$和$x=\frac{\sqrt{3}}{3}$是$f(x)$的極值點(diǎn)。7.（1）$f(x)$在$(-\infty,0)$上單調(diào)遞增解析：函數(shù)$f(x)=\frac{x}{x^2+1}$的導(dǎo)數(shù)為$f'(x)=\frac{1}{(x^2+1)^2}$，由于$(x^2+1)^2>0$，所以$f'(x)>0$，函數(shù)在$(-\infty,0)$上單調(diào)遞增。8.（1）$f(x)$在$(-\infty,0)$上單調(diào)遞增解析：函數(shù)$f(x)=\log_2(x+1)$的導(dǎo)數(shù)為$f'(x)=\frac{1}{(x+1)\ln2}$，由于$x+1>0$，$\ln2>0$，所以$f'(x)>0$，函數(shù)在$(-\infty,0)$上單調(diào)遞增。9.（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增解析：函數(shù)$f(x)=\sqrt{x^2+1}$的導(dǎo)數(shù)為$f'(x)=\frac{x}{\sqrt{x^2+1}}$，當(dāng)$x>0$或$x<0$時(shí)，$f'(x)>0$，函數(shù)在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增。10.（3）$f(x)$在$(-\infty,0)\cup(0,+\infty)$上單調(diào)遞增解析：函數(shù)$f(x)=x^3-x$的導(dǎo)數(shù)為$f'(