Bisection method for approximate roots of equations Understand the relation between a function's zeros and an equation's solutions, the zero-existence theorem, and use bisection to find approximate roots.
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 f（x）=㏑x+2x-5的零点一定位于以下的区间（ ） A. （1，2） B. （2，3） C. （3，4） D. （4，5） 3 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 ) 4 已知曲线y = ∣ x ∣ e x y=\dfrac{|x|}{{{e}^{x}}} y = e x ∣ x ∣ x = − 1 x=-1 x = − 1 x = x 0 ( x 0 > 0 ) x={{x}_{0}}\left({x}_{0} > 0\right) x = x 0 ( x 0 > 0 ) x 0 ∈ ( m 4 , m + 1 4 ) , m ∈ Z {{x}_{0}}\in (\dfrac{m}{4},\dfrac{m+1}{4}),m\in Z x 0 ∈ ( 4 m , 4 m + 1 ) , m ∈ Z 5 已知函数f ( x ) = e x − k − x f(x)=e^{x-k}-x f ( x ) = e x − k − x ( x ∈ R ) ( 1 ) (x∈R)(1) ( x ∈ R ) ( 1 ) k = 0 k=0 k = 0 f ( x ) ⩾ m f(x)\geqslant m f ( x ) ⩾ m R R R m m m k > 1 k > 1 k > 1 f ( x ) f(x) f ( x ) 6 方程log 2 x + x − 5 = 0 \log _{2}x+x-5=0 log 2 x + x − 5 = 0 ( ) () ( ) ( 1 , 2 ) (1,2) ( 1 , 2 ) ( 2 , 3 ) (2,3) ( 2 , 3 ) ( 3 , 4 ) (3,4) ( 3 , 4 ) ( 4 , 5 ) (4,5) ( 4 , 5 ) 7 已知函数f ( x ) = ( x − 1 ) ln x f\left(x\right)=\left(x-1\right)\ln x f ( x ) = ( x − 1 ) ln x . g ( x ) = x − ln x − 3 e ( 1 ) .g\left(x\right)=x-\ln x-\dfrac{3}{e}(1) . g ( x ) = x − ln x − e 3 ( 1 ) y = f ( x ) y=f\left(x\right) y = f ( x ) 8 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 ) 9 已知函数 F ( x ) = p x + ln ( p x ) ( F(x)= \dfrac{p}{x}+\ln (px) ( F ( x ) = x p + ln ( p x ) ( p > 0 ) . ( 1 ) p > 0).(1) p > 0 ) . ( 1 ) p = 1 2 p=\dfrac{1}{2} p = 2 1 F ( x ) F(x) F ( x ) k k k 10 设函数f ( x ) = x 2 2 − k ln x f(x)= \dfrac {x^{2}}{2}-k\ln x f ( x ) = 2 x 2 − k ln x k ∈ R k∈R k ∈ R f ( x ) f(x) f ( x ) f ( x ) = 0 f(x)=0 f ( x ) = 0 ( 1 , e ) (1, \sqrt {e}) ( 1 , e )