Skip to main content
Library homepage
 
Global

4.3: 评估函数

计算函数时,将 x 替换为给定的数值或代数表达式,然后简化结果。

假定f(x)=x2+5x+12,找到以下各项:

  1. f(2)
  2. f(5)
  3. f(t)
  4. f(4x1)
解决方案
  1. 替换x2

f(x)=x2+5x+12Given equationf(2)=(22)+52+12Replace x with 2 - notice that only 2 is squared, not the minus signf(2)=4+10+12Simplifyf(2)=18Solution

  1. 替换x5

f(x)=x2+5x+12Given equation f(5)=((5)2)+5(5)+12Replace x with −5 - notice that it’s x2 , so the −5 is squared, but the result is still negativef(2)=25+(25)+12Simplify f(2)=38Solution

  1. 替换xt

f(x)=x2+5x+12Given equation f(t)=t2+5(t)+12Replace x with t f(t)=t2+5t+12Simplify 

  1. 替换x(4x1)

f(x)=x2+5x+12Given equationf(4x1)=(4x1)2+5(4x1)+12Replace x with (4x1)f(4x1)=(16x28x+1)+20x5+12Expand(4x1)2 and distribute the 5 to (4x1)f(4x1)=16x2+8x1+20x5+12Distribute the negative signf(4x1)=16x2+28x+6Simplify

  1. 对于该函数f(x)=x39,请找到
    1. f(3)
    2. f(2x5)
    3. f(t)
  2. 对于该函数f(x)=x24x+1,请找到
    1. f(2)
    2. f(4x+3)
    3. f(z)
  3. 对于该函数f(x)=x4+9x26x1,请找到
    1. f(1)
    2. f(x+2)
    3. f(a)