4.3: Evaluar una función
- Page ID
- 112501
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Cuando se evalúa una función, reemplace x con un valor numérico dado o una expresión algebraica, y luego simplifique el resultado.
Dado\(f(x) = −x^2 + 5x + 12\), encuentra cada uno de los siguientes:
- \(f(2)\)
- \(f(−5)\)
- \(f(t)\)
- \(f(4x − 1)\)
Solución
- Reemplazar\(x\) con\(2\):
\(\begin{aligned} f(x)& = −x^ 2 + 5x + 12 && \text{Given equation} \\f(2) &= −(2^2 ) + 5 ∗ 2 + 12 && \text{Replace x with 2 - notice that only 2 is squared, not the minus sign} \\ f(2) &= −4 + 10 + 12 &&\text{Simplify} \\ f(2)& = 18 &&\text{Solution}\end{aligned}\)
- Reemplazar\(x\) con\(-5\):
\(\begin{aligned} f(x) &= −x ^2 + 5x + 12 &&\text{Given equation } \\ f(−5) &= −((−5)^2 ) + 5 ∗ (−5) + 12 &&\text{Replace x with −5 - notice that it’s }−x ^2 \text{ , so the −5 is squared, but the result is still negative} \\ f(2) &= −25 + (−25) + 12 && \text{Simplify } \\ f(2) &= −38 &&\text{Solution} \end{aligned}\)
- Reemplazar\(x\) con\(t\):
\(\begin{aligned} f(x) &= −x^2 + 5x + 12 &&\text{Given equation } \\ f(t)& = −t ^2 + 5 ∗ (t) + 12 && \text{Replace x with t } \\ f(t) &= −t 2 + 5t + 12&& \text{Simplify } \end{aligned}\)
- Reemplazar\(x\) con\((4x-1)\):
\(\begin{aligned} f(x) &= −x^2 + 5x + 12&& \text{Given equation} \\ f(4x − 1) &= −(4x − 1)^2 + 5 ∗ (4x − 1) + 12 &&\text{Replace x with }(4x - 1) \\ f(4x − 1) &= −(16x ^2 − 8x + 1) + 20x − 5 + 12 &&\text{Expand} (4x − 1)^2 \text{ and distribute the }5 \text{ to } (4x - 1) \\ f(4x − 1) &= −16x^ 2 + 8x − 1 + 20x − 5 + 12 && \text{Distribute the negative sign}\\ f(4x − 1) &= 16x ^2 + 28x + 6 && \text{Simplify} \end{aligned}\)
- Para la función\(f(x) = x^3 − 9\), encontrar
- \(f(3)\)
- \(f(2x-5)\)
- \(f(t)\)
- Para la función\(f(x) = x^2 − 4x + 1\), encontrar
- \(f(2)\)
- \(f(4x+3)\)
- \(f(z)\)
- Para la función\(f(x) = x^4 + 9x^2 − 6x − 1\), encontrar
- \(f(1)\)
- \(f(x+2)\)
- \(f(a)\)