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Query

4.3: Kutathmini Kazi

  • Page ID
    164613
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    Wakati kazi ni tathmini, kuchukua nafasi ya x na kupewa thamani numeric au kujieleza algebraic, na kisha kurahisisha matokeo.

    Kutokana\(f(x) = −x^2 + 5x + 12\), tafuta kila moja ya yafuatayo:

    1. \(f(2)\)
    2. \(f(−5)\)
    3. \(f(t)\)
    4. \(f(4x − 1)\)
    Suluhisho
    1. Badilisha nafasi\(x\) na\(2\):

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

    1. Badilisha nafasi\(x\) na\(-5\):

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

    1. Badilisha nafasi\(x\) na\(t\):

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

    1. Badilisha nafasi\(x\) na\((4x-1)\):

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

    1. Kwa kazi\(f(x) = x^3 − 9\), tafuta
      1. \(f(3)\)
      2. \(f(2x-5)\)
      3. \(f(t)\)
    2. Kwa kazi\(f(x) = x^2 − 4x + 1\), tafuta
      1. \(f(2)\)
      2. \(f(4x+3)\)
      3. \(f(z)\)
    3. Kwa kazi\(f(x) = x^4 + 9x^2 − 6x − 1\), tafuta
      1. \(f(1)\)
      2. \(f(x+2)\)
      3. \(f(a)\)