Concept and meaning of the derivative Move from average to instantaneous rate of change, understand the derivative as instantaneous rate of change, the limit idea, and its geometric meaning.
Practice questions 1 设f ( x ) f(x) f ( x ) [ a , + ∞ ) [a,+∞) [ a , + ∞ ) f ′ ( x ) f'(x) f ′ ( x ) x > a x>a x > a f ′ ( x ) > k > 0 ( k f'(x)>k>0(k f ′ ( x ) > k > 0 ( k ) ) ) f ( a ) < 0 f(a)< 0 f ( a ) < 0 ( a , a − f ( a ) k ) (a,a-\dfrac{f(a)}{k}) ( a , a − k f ( a ) ) f ( x ) = 0 f(x)=0 f ( x ) = 0 2 已知a a a x x x y y y ( 2 x − y e ) ⋅ ln y x = x a e \left(2x- \dfrac{y}{e}\right)·\ln \dfrac{y}{x}= \dfrac{x}{ae} ( 2 x − e y ) ⋅ ln x y = a e x a a a e e e 3 已知0 < a < 5 0< a< 5 0 < a < 5 a ln 5 = 5 ln a a\ln 5=5\ln a a ln 5 = 5 ln a 0 < b < 6 0< b< 6 0 < b < 6 b ln 6 = 6 ln b b\ln 6=6\ln b b ln 6 = 6 ln b 0 < c < 7 0< c< 7 0 < c < 7 c ln 7 = 7 ln c c\ln 7=7\ln c c ln 7 = 7 ln c a a a b b b c c c ( ) (\quad) ( ) 4 已知圆C 1 C_{1} C 1 x 2 + y 2 = r 2 ( r > 0 ) x^{2}+y^{2}=r^{2}(r > 0) x 2 + y 2 = r 2 ( r > 0 ) C 2 C_{2} C 2 x 2 = 2 p y ( p > 0 ) x^{2}=2py(p > 0) x 2 = 2 p y ( p > 0 ) ( 2 , − 2 ) ( \sqrt {2},-2) ( 2 , − 2 ) C 1 C_{1} C 1 C 2 C_{2} C 2 5 若函数f ( x ) f\left( x \right) f ( x ) D D D I I I F ( x ) = f ( x ) x F\left( x \right)=\dfrac{f\left( x \right)}{x} F ( x ) = x f ( x ) I I I y = f ( x ) y=f\left( x \right) y = f ( x ) 6 设曲线f ( x ) = a x 2 f(x)=ax^{2} f ( x ) = a x 2 ( 2 , 4 a ) (2,4a) ( 2 , 4 a ) 4 x − y + 4 = 0 4x-y+4=0 4 x − y + 4 = 0 a = ( ) a=(\quad) a = ( ) 2 2 2 − 1 16 -\dfrac{1}{16} − 16 1 1 2 \dfrac{1}{2} 2 1 7 已知函数f ( x ) = a ln ( x + 1 ) − x 2 f(x)=a \ln (x+1)-x^{2} f ( x ) = a ln ( x + 1 ) − x 2 ( 0 , 1 ) (0,1) ( 0 , 1 ) p p p q q q p ≠ q p\neq q p = q f ( p + 1 ) − f ( q + 1 ) p − q > 1 \dfrac{f\left( p+1 \right)-f\left( q+1 \right)}{p-q} > 1 p − q f ( p + 1 ) − f ( q + 1 ) > 1 8 已知函数f ( x ) = − x 4 + 8 x 2 + 2021 f(x)=-x^{4}+8x^{2}+2021 f ( x ) = − x 4 + 8 x 2 + 2021 f ( x ) f(x) f ( x ) ( ) (\quad) ( ) ( − 3 , 0 ) (-3,0) ( − 3 , 0 ) ( 0 , 1 ) (0,1) ( 0 , 1 ) ( 1 , 3 ) (1,3) ( 1 , 3 ) ( 2 , 4 ) (2,4) ( 2 , 4 ) 9 若x = − 4 x=-4 x = − 4 f ( x ) = x 2 ( x + 2 m ) f(x)=x^{2}(x+2m) f ( x ) = x 2 ( x + 2 m ) f ( x ) f(x) f ( x ) 10 已知函数f ( x ) = − 1 8 x + 3 cos 2 x f(x)=-\dfrac{1}{8}x+3\cos 2x f ( x ) = − 8 1 x + 3 cos 2 x x = 0 x=0 x = 0 y = n x y=nx y = n x ( 1 + x + x 2 ) ( 1 + x ) n = a 0 + a 1 x + a 2 x 2 + ⋯ + a n + 2 x n + 2 (1+x+x^{2})(1+x)^{n}=a_{0}+a_{1}x+a_{2}x^{2}+⋯+a_{n+2}x^{n+2} ( 1 + x + x 2 ) ( 1 + x ) n = a 0 + a 1 x + a 2 x 2 + ⋯ + a n + 2 x n + 2 11 若f ( x ) = ln x f(x)=\ln x f ( x ) = ln x △ x → 0 lim f ( 1 + 2 △ x ) − f ( 1 ) △ x = ( ) \overset{\lim}{\triangle x\rightarrow 0}\dfrac{f(1+2\triangle x)-f(1)}{\triangle x}=(\quad) △ x → 0 l i m △ x f ( 1 + 2△ x ) − f ( 1 ) = ( ) 1 1 1 12 已知函数f ( x ) = x sin x + a cos x + x f(x)=x\sin x+a\cos x+x f ( x ) = x sin x + a cos x + x a ∈ R a∈R a ∈ R ( ( ( ) ) ) a = − 1 a=-1 a = − 1 y = f ( x ) y=f(x) y = f ( x ) ( 0 , f ( 0 ) ) (0,f(0)) ( 0 , f ( 0 )) ( ( ( ) ) ) a = 2 a=2 a = 2 f ( x ) f(x) f ( x ) 13 对于R上可导的任意函数f（x），若满足（x-3）f′（x）≥0，则必有（ ） A. f（0）+f（5）＜2f（3） B. f（0）+f（5）≤2f（3） C. f（0）+f（5）≥2f（3） D. f（0）+f（5）＞2f（3） 14 已知函数f ( x ) = ln x + ( a − 1 ) x + 3 x − 6 − 2 a f(x)=\ln x+(a-1)x+\dfrac{3}{x}-6-2a f ( x ) = ln x + ( a − 1 ) x + x 3 − 6 − 2 a ( 2 , 4 ) (2,4) ( 2 , 4 ) a a a 15 若曲线f ( x ) = e x + m x f(x)=e^{x}+\dfrac{m}{x} f ( x ) = e x + x m ( − ∞ , 0 ) (-∞,0) ( − ∞ , 0 ) y y y m m m ( ) (\quad) ( ) ( -∞, 4 e 2 ] (\text{-∞,}\dfrac{4}{e^{2}}\text{]} ( -∞, e 2 4 ] 16 已知函数f ( x ) = x + a 2 x . f(x)=x+\dfrac{a}{2x}. f ( x ) = x + 2 x a . y = f ( x ) y=f(x) y = f ( x ) ( 2 , 0 ) \left({2,0}\right) ( 2 , 0 ) a a a ( ) (\quad) ( ) ( − ∞ , 1 ) ∪ ( 8 , + ∞ ) (-∞,1)∪(8,+∞) ( − ∞ , 1 ) ∪ ( 8 , + ∞ ) 17 已知定义在R上的奇函数f（x），f（1）=2，对任意的x＞0，f'（x）＞2，则不等式f（x）＞2x的解集为（ ） A. （-∞，-1）∪（0，1） B. （-1，0）∪（0，1） C. （-∞，-1）∪（1，+∞） D. （-1，0）∪（… 18 已知( x + 4 x − 4 ) 6 = a 1 x 6 + a 2 x 5 + ⋯ + a 7 + ⋯ + a 13 x − 6 (x+\dfrac{4}{x}-4)^{6}=a_{1}x^{6}+a_{2}x^{5}+⋯+a_{7}+⋯+a_{13}x^{-6} ( x + x 4 − 4 ) 6 = a 1 x 6 + a 2 x 5 + ⋯ + a 7 + ⋯ + a 13 x − 6 a 2 = a_{2}= a 2 = 6 a 1 + 5 a 2 + ⋯ + a 6 − a 8 − 2 a 9 − ⋯ − 6 a 13 = 6a_{1}+5a_{2}+⋯+a_{6}-a_{8}-2a_{9}-⋯-6a_{13}= 6 a 1 + 5 a 2 + ⋯ + a 6 − a 8 − 2 a 9 − ⋯ − 6 a 13 = 19 函数f ( x ) = 1 + ln x x f(x)= \dfrac {1+\ln x}{x} f ( x ) = x 1 + ln x ( ( ( ) ) ) f ( x ) f(x) f ( x ) f ( x ) f(x) f ( x ) ( ( ( ) ) ) x ⩾ 1 x\geqslant 1 x ⩾ 1 ( x + 1 ) ( 1 + ln x ) x ⩾ 2 \dfrac {(x+1)(1+\ln x)}{x}\geqslant 2 x ( x + 1 ) ( 1 + ln x ) ⩾ 2 20 若函数f ( x ) = x 2 − 2 x + a ln x f(x)=x^{2}-2x+a\ln x f ( x ) = x 2 − 2 x + a ln x a a a ( ) (\quad) ( ) a > 1 2 a>\dfrac{1}{2} a > 2 1 − 1 2 < a < 0 -\dfrac{1}{2}< a< 0 − 2 1 < a < 0 a < 1 2 a< \dfrac {1}{2} a < 2 1