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。

示例 Template:Index

假定$f(x) = −x^2 + 5x + 12$，找到以下各项：

1. $f(2)$
2. $f(−5)$
3. $f(t)$
4. $f(4x − 1)$

解决方案

1. 替换$x$为$2$：

$\begin{aligned} f(x)& = −x^ 2 + 5x + 12 && \text{Given equation} \\f(2) &= −(2^2 ) + 5 ∗ 2 + 12 && \text{Replace x with 2 - notice that only 2 is squared, not the minus sign} \\ f(2) &= −4 + 10 + 12 &&\text{Simplify} \\ f(2)& = 18 &&\text{Solution}\end{aligned}$

2. 替换$x$为$-5$：

$\begin{aligned} f(x) &= −x ^2 + 5x + 12 &&\text{Given equation } \\ f(−5) &= −((−5)^2 ) + 5 ∗ (−5) + 12 &&\text{Replace x with −5 - notice that it’s }−x ^2 \text{ , so the −5 is squared, but the result is still negative} \\ f(2) &= −25 + (−25) + 12 && \text{Simplify } \\ f(2) &= −38 &&\text{Solution} \end{aligned}$

3. 替换$x$为$t$：

$\begin{aligned} f(x) &= −x^2 + 5x + 12 &&\text{Given equation } \\ f(t)& = −t ^2 + 5 ∗ (t) + 12 && \text{Replace x with t } \\ f(t) &= −t 2 + 5t + 12&& \text{Simplify } \end{aligned}$

4. 替换$x$为$(4x-1)$：

$\begin{aligned} f(x) &= −x^2 + 5x + 12&& \text{Given equation} \\ f(4x − 1) &= −(4x − 1)^2 + 5 ∗ (4x − 1) + 12 &&\text{Replace x with }(4x - 1) \\ f(4x − 1) &= −(16x ^2 − 8x + 1) + 20x − 5 + 12 &&\text{Expand} (4x − 1)^2 \text{ and distribute the }5 \text{ to } (4x - 1) \\ f(4x − 1) &= −16x^ 2 + 8x − 1 + 20x − 5 + 12 && \text{Distribute the negative sign}\\ f(4x − 1) &= 16x ^2 + 28x + 6 && \text{Simplify} \end{aligned}$

练习 Template:Index

1. 对于该函数$f(x) = x^3 − 9$，请找到
   1. $f(3)$
   2. $f(2x-5)$
   3. $f(t)$
2. 对于该函数$f(x) = x^2 − 4x + 1$，请找到
   1. $f(2)$
   2. $f(4x+3)$
   3. $f(z)$
3. 对于该函数$f(x) = x^4 + 9x^2 − 6x − 1$，请找到
   1. $f(1)$
   2. $f(x+2)$
   3. $f(a)$