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Matemáticas 4 · Tema 1

Evaluación de funciones

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Apuntes

La x pertenece al dominio de la función si la regla de correspondencia está bien definida y al evaluarla nos arroja un solo valor.

f(x)=3x−1f(x) = \dfrac{3}{x-1}

La parte de la derecha es la regla de correspondencia: la fórmula que da la imagen. Evaluar es sustituir el valor en lugar de las xx y resolver de manera normal.

Ejemplo 1

Dada la función f(x)=−5x2−x+3f(x) = -5x^2 - x + 3, encuentra f(−2)f(-2).

f(−2)=−5(−2)2−(−2)+3=−5(4)+2+3=−20+5=−15\begin{aligned} f(-2) &= -5(-2)^2 - (-2) + 3 \\ &= -5(4) + 2 + 3 \\ &= -20 + 5 = \mathbf{-15} \end{aligned}

La regla de correspondencia está bien definida y da un solo valor: x=−2x = -2 sí es parte del dominio.

Ejemplo 2

dadas las funciones

f(x)=3x2−5x+2f(x) = 3x^2 - 5x + 2
m(x)=34x2−1m(x) = \dfrac{3}{4x^2-1}
t(x)=2x+3t(x) = 2x + 3
L(x)=3x−24−5xL(x) = \dfrac{3x-2}{4-5x}
b(x)=−5x2b(x) = -5x^2
g(x)=−2x3+3x+1g(x) = -2x^3 + 3x + 1

encuentra

f(−4)f(-4) g(−1)g(-1) m(2)m(2) L(3)L(3) b(y3)b(y^3) g(2a4)g(2a^4) b(a+h)b(a+h) t(a−h)t(a-h)
f(−4)=3(−4)2−5(−4)+2=3(16)+20+2=48+22=70\begin{aligned} f(-4) &= 3(-4)^2 - 5(-4) + 2 \\ &= 3(16) + 20 + 2 \\ &= 48 + 22 = \mathbf{70} \end{aligned}
g(−1)=−2(−1)3+3(−1)+1=−2(−1)−3+1=2−3+1=0\begin{aligned} g(-1) &= -2(-1)^3 + 3(-1) + 1 \\ &= -2(-1) - 3 + 1 \\ &= 2 - 3 + 1 = \mathbf{0} \end{aligned}
m(2)=34(2)2−1=34(4)−1=315=15m(2) = \dfrac{3}{4(2)^2-1} = \dfrac{3}{4(4)-1} = \dfrac{3}{15} = \mathbf{\dfrac{1}{5}}
L(3)=3(3)−24−5(3)=9−24−15=7−11=−711L(3) = \dfrac{3(3)-2}{4-5(3)} = \dfrac{9-2}{4-15} = \dfrac{7}{-11} = \mathbf{-\dfrac{7}{11}}
b(y3)=−5(y3)2=−5y6b(y^3) = -5(y^3)^2 = \mathbf{-5y^6}
g(2a4)=−2(2a4)3+3(2a4)+1=−2(8a12)+6a4+1=−16a12+6a4+1\begin{aligned} g(2a^4) &= -2(2a^4)^3 + 3(2a^4) + 1 \\ &= -2(8a^{12}) + 6a^4 + 1 \\ &= \mathbf{-16a^{12} + 6a^4 + 1} \end{aligned}
b(a+h)=−5(a+h)2=−5(a2+2ah+h2)=−5a2−10ah−5h2\begin{aligned} b(a+h) &= -5(a+h)^2 = -5(a^2 + 2ah + h^2) \\ &= \mathbf{-5a^2 - 10ah - 5h^2} \end{aligned}
t(a−h)=2(a−h)+3=2a−2h+3t(a-h) = 2(a-h) + 3 = \mathbf{2a - 2h + 3}

Ejercicios

Bloque 1

dadas las funciones

b(x)=−3x2b(x) = -3x^2
f(x)=4x2−3x+1f(x) = 4x^2 - 3x + 1
t(x)=5x−2t(x) = 5x - 2
m(x)=53x2−1m(x) = \dfrac{5}{3x^2-1}
g(x)=−2x3+4x−1g(x) = -2x^3 + 4x - 1
L(x)=4x−23−5xL(x) = \dfrac{4x-2}{3-5x}

encuentra

g(−2)g(-2) f(−1)f(-1) L(2)L(2) m(3)m(3) b(y4)b(y^4) g(−3x2)g(-3x^2) b(a+h)b(a+h) t(a−h)t(a-h)
Bloque 2

dadas las funciones

b(x)=5x2b(x) = 5x^2
f(x)=2x2+5x+3f(x) = 2x^2 + 5x + 3
t(x)=3x+4t(x) = 3x + 4
m(x)=25x2−4m(x) = \dfrac{2}{5x^2-4}
g(x)=3x3+2x−1g(x) = 3x^3 + 2x - 1
L(x)=5x−16+2xL(x) = \dfrac{5x-1}{6+2x}

encuentra

g(3)g(3) f(3)f(3) L(−4)L(-4) m(−2)m(-2) b(2y5)b(2y^5) g(4x2)g(4x^2) b(a+h)b(a+h) t(a−h)t(a-h)
Bloque 3

dadas las funciones

b(x)=−2x2b(x) = -2x^2
f(x)=−4x2−3x+5f(x) = -4x^2 - 3x + 5
t(x)=2x−1t(x) = 2x - 1
m(x)=13+2x2m(x) = \dfrac{1}{3+2x^2}
g(x)=−3x3−4x−2g(x) = -3x^3 - 4x - 2
L(x)=3x−24−2xL(x) = \dfrac{3x-2}{4-2x}

encuentra

g(1)g(1) f(−5)f(-5) L(−1)L(-1) m(4)m(4) b(−y4)b(-y^4) g(−x2)g(-x^2) b(a+h)b(a+h) t(a−h)t(a-h)
Bloque 4

dadas las funciones

b(x)=4x2b(x) = 4x^2
f(x)=−3x2+2x−6f(x) = -3x^2 + 2x - 6
t(x)=6x+1t(x) = 6x + 1
m(x)=43−4x2m(x) = \dfrac{4}{3-4x^2}
g(x)=−5x3+2x+6g(x) = -5x^3 + 2x + 6
L(x)=−2x−15−xL(x) = \dfrac{-2x-1}{5-x}

encuentra

g(2)g(2) f(2)f(2) L(3)L(3) m(−1)m(-1) b(3y4)b(3y^4) g(2x2)g(2x^2) b(a+h)b(a+h) t(a−h)t(a-h)
Bloque 5

dadas las funciones

b(x)=6x2b(x) = 6x^2
f(x)=7x2−x+4f(x) = 7x^2 - x + 4
t(x)=−3x+2t(x) = -3x + 2
m(x)=56x2−1m(x) = \dfrac{5}{6x^2-1}
g(x)=−x3−x+1g(x) = -x^3 - x + 1
L(x)=5x−44+3xL(x) = \dfrac{5x-4}{4+3x}

encuentra

g(−5)g(-5) f(−3)f(-3) L(2)L(2) m(3)m(3) b(−2y5)b(-2y^5) g(−4x2)g(-4x^2) b(a+h)b(a+h) t(a−h)t(a-h)
Bloque 6

dadas las funciones

b(x)=−4x2b(x) = -4x^2
f(x)=5x2−3x−1f(x) = 5x^2 - 3x - 1
t(x)=7x−4t(x) = 7x - 4
m(x)=43x2−3m(x) = \dfrac{4}{3x^2-3}
g(x)=5x3−4x−1g(x) = 5x^3 - 4x - 1
L(x)=−3x+26−3xL(x) = \dfrac{-3x+2}{6-3x}

encuentra

g(4)g(4) f(5)f(5) L(−5)L(-5) m(5)m(5) b(6y3)b(6y^3) g(3x2)g(3x^2) b(a+h)b(a+h) t(a−h)t(a-h)

Respuestas

▼
Bloque 1
g(−2)=−2(−2)3+4(−2)−1=−2(−8)−8−1=16−9=7\begin{aligned} g(-2) &= -2(-2)^3 + 4(-2) - 1 \\ &= -2(-8) - 8 - 1 \\ &= 16 - 9 = \mathbf{7} \end{aligned}
f(−1)=4(−1)2−3(−1)+1=4(1)+3+1=8\begin{aligned} f(-1) &= 4(-1)^2 - 3(-1) + 1 \\ &= 4(1) + 3 + 1 = \mathbf{8} \end{aligned}
L(2)=4(2)−23−5(2)=8−23−10=6−7=−67L(2) = \dfrac{4(2)-2}{3-5(2)} = \dfrac{8-2}{3-10} = \dfrac{6}{-7} = \mathbf{-\dfrac{6}{7}}
m(3)=53(3)2−1=53(9)−1=526m(3) = \dfrac{5}{3(3)^2-1} = \dfrac{5}{3(9)-1} = \mathbf{\dfrac{5}{26}}
b(y4)=−3(y4)2=−3y8b(y^4) = -3(y^4)^2 = \mathbf{-3y^8}
g(−3x2)=−2(−3x2)3+4(−3x2)−1=−2(−27x6)−12x2−1=54x6−12x2−1\begin{aligned} g(-3x^2) &= -2(-3x^2)^3 + 4(-3x^2) - 1 \\ &= -2(-27x^6) - 12x^2 - 1 \\ &= \mathbf{54x^6 - 12x^2 - 1} \end{aligned}
b(a+h)=−3(a+h)2=−3(a2+2ah+h2)=−3a2−6ah−3h2\begin{aligned} b(a+h) &= -3(a+h)^2 = -3(a^2 + 2ah + h^2) \\ &= \mathbf{-3a^2 - 6ah - 3h^2} \end{aligned}
t(a−h)=5(a−h)−2=5a−5h−2t(a-h) = 5(a-h) - 2 = \mathbf{5a - 5h - 2}
Bloque 2
g(3)=3(3)3+2(3)−1=3(27)+6−1=81+5=86\begin{aligned} g(3) &= 3(3)^3 + 2(3) - 1 \\ &= 3(27) + 6 - 1 \\ &= 81 + 5 = \mathbf{86} \end{aligned}
f(3)=2(3)2+5(3)+3=2(9)+15+3=18+18=36\begin{aligned} f(3) &= 2(3)^2 + 5(3) + 3 \\ &= 2(9) + 15 + 3 \\ &= 18 + 18 = \mathbf{36} \end{aligned}
L(−4)=5(−4)−16+2(−4)=−20−16−8=−21−2=212L(-4) = \dfrac{5(-4)-1}{6+2(-4)} = \dfrac{-20-1}{6-8} = \dfrac{-21}{-2} = \mathbf{\dfrac{21}{2}}
m(−2)=25(−2)2−4=25(4)−4=216=18m(-2) = \dfrac{2}{5(-2)^2-4} = \dfrac{2}{5(4)-4} = \dfrac{2}{16} = \mathbf{\dfrac{1}{8}}
b(2y5)=5(2y5)2=5(4y10)=20y10b(2y^5) = 5(2y^5)^2 = 5(4y^{10}) = \mathbf{20y^{10}}
g(4x2)=3(4x2)3+2(4x2)−1=3(64x6)+8x2−1=192x6+8x2−1\begin{aligned} g(4x^2) &= 3(4x^2)^3 + 2(4x^2) - 1 \\ &= 3(64x^6) + 8x^2 - 1 \\ &= \mathbf{192x^6 + 8x^2 - 1} \end{aligned}
b(a+h)=5(a+h)2=5(a2+2ah+h2)=5a2+10ah+5h2\begin{aligned} b(a+h) &= 5(a+h)^2 = 5(a^2 + 2ah + h^2) \\ &= \mathbf{5a^2 + 10ah + 5h^2} \end{aligned}
t(a−h)=3(a−h)+4=3a−3h+4t(a-h) = 3(a-h) + 4 = \mathbf{3a - 3h + 4}
Bloque 3
g(1)=−3(1)3−4(1)−2=−3−4−2=−9\begin{aligned} g(1) &= -3(1)^3 - 4(1) - 2 \\ &= -3 - 4 - 2 = \mathbf{-9} \end{aligned}
f(−5)=−4(−5)2−3(−5)+5=−4(25)+15+5=−100+20=−80\begin{aligned} f(-5) &= -4(-5)^2 - 3(-5) + 5 \\ &= -4(25) + 15 + 5 \\ &= -100 + 20 = \mathbf{-80} \end{aligned}
L(−1)=3(−1)−24−2(−1)=−3−24+2=−56L(-1) = \dfrac{3(-1)-2}{4-2(-1)} = \dfrac{-3-2}{4+2} = \mathbf{-\dfrac{5}{6}}
m(4)=13+2(4)2=13+2(16)=135m(4) = \dfrac{1}{3+2(4)^2} = \dfrac{1}{3+2(16)} = \mathbf{\dfrac{1}{35}}
b(−y4)=−2(−y4)2=−2y8b(-y^4) = -2(-y^4)^2 = \mathbf{-2y^8}
g(−x2)=−3(−x2)3−4(−x2)−2=−3(−x6)+4x2−2=3x6+4x2−2\begin{aligned} g(-x^2) &= -3(-x^2)^3 - 4(-x^2) - 2 \\ &= -3(-x^6) + 4x^2 - 2 \\ &= \mathbf{3x^6 + 4x^2 - 2} \end{aligned}
b(a+h)=−2(a+h)2=−2(a2+2ah+h2)=−2a2−4ah−2h2\begin{aligned} b(a+h) &= -2(a+h)^2 = -2(a^2 + 2ah + h^2) \\ &= \mathbf{-2a^2 - 4ah - 2h^2} \end{aligned}
t(a−h)=2(a−h)−1=2a−2h−1t(a-h) = 2(a-h) - 1 = \mathbf{2a - 2h - 1}
Bloque 4
g(2)=−5(2)3+2(2)+6=−5(8)+4+6=−40+10=−30\begin{aligned} g(2) &= -5(2)^3 + 2(2) + 6 \\ &= -5(8) + 4 + 6 \\ &= -40 + 10 = \mathbf{-30} \end{aligned}
f(2)=−3(2)2+2(2)−6=−3(4)+4−6=−12−2=−14\begin{aligned} f(2) &= -3(2)^2 + 2(2) - 6 \\ &= -3(4) + 4 - 6 \\ &= -12 - 2 = \mathbf{-14} \end{aligned}
L(3)=−2(3)−15−3=−6−12=−72L(3) = \dfrac{-2(3)-1}{5-3} = \dfrac{-6-1}{2} = \mathbf{-\dfrac{7}{2}}
m(−1)=43−4(−1)2=43−4(1)=4−1=−4m(-1) = \dfrac{4}{3-4(-1)^2} = \dfrac{4}{3-4(1)} = \dfrac{4}{-1} = \mathbf{-4}
b(3y4)=4(3y4)2=4(9y8)=36y8b(3y^4) = 4(3y^4)^2 = 4(9y^8) = \mathbf{36y^8}
g(2x2)=−5(2x2)3+2(2x2)+6=−5(8x6)+4x2+6=−40x6+4x2+6\begin{aligned} g(2x^2) &= -5(2x^2)^3 + 2(2x^2) + 6 \\ &= -5(8x^6) + 4x^2 + 6 \\ &= \mathbf{-40x^6 + 4x^2 + 6} \end{aligned}
b(a+h)=4(a+h)2=4(a2+2ah+h2)=4a2+8ah+4h2\begin{aligned} b(a+h) &= 4(a+h)^2 = 4(a^2 + 2ah + h^2) \\ &= \mathbf{4a^2 + 8ah + 4h^2} \end{aligned}
t(a−h)=6(a−h)+1=6a−6h+1t(a-h) = 6(a-h) + 1 = \mathbf{6a - 6h + 1}
Bloque 5
g(−5)=−(−5)3−(−5)+1=−(−125)+5+1=125+6=131\begin{aligned} g(-5) &= -(-5)^3 - (-5) + 1 \\ &= -(-125) + 5 + 1 \\ &= 125 + 6 = \mathbf{131} \end{aligned}
f(−3)=7(−3)2−(−3)+4=7(9)+3+4=63+7=70\begin{aligned} f(-3) &= 7(-3)^2 - (-3) + 4 \\ &= 7(9) + 3 + 4 \\ &= 63 + 7 = \mathbf{70} \end{aligned}
L(2)=5(2)−44+3(2)=10−44+6=610=35L(2) = \dfrac{5(2)-4}{4+3(2)} = \dfrac{10-4}{4+6} = \dfrac{6}{10} = \mathbf{\dfrac{3}{5}}
m(3)=56(3)2−1=56(9)−1=553m(3) = \dfrac{5}{6(3)^2-1} = \dfrac{5}{6(9)-1} = \mathbf{\dfrac{5}{53}}
b(−2y5)=6(−2y5)2=6(4y10)=24y10b(-2y^5) = 6(-2y^5)^2 = 6(4y^{10}) = \mathbf{24y^{10}}
g(−4x2)=−(−4x2)3−(−4x2)+1=−(−64x6)+4x2+1=64x6+4x2+1\begin{aligned} g(-4x^2) &= -(-4x^2)^3 - (-4x^2) + 1 \\ &= -(-64x^6) + 4x^2 + 1 \\ &= \mathbf{64x^6 + 4x^2 + 1} \end{aligned}
b(a+h)=6(a+h)2=6(a2+2ah+h2)=6a2+12ah+6h2\begin{aligned} b(a+h) &= 6(a+h)^2 = 6(a^2 + 2ah + h^2) \\ &= \mathbf{6a^2 + 12ah + 6h^2} \end{aligned}
t(a−h)=−3(a−h)+2=−3a+3h+2t(a-h) = -3(a-h) + 2 = \mathbf{-3a + 3h + 2}
Bloque 6
g(4)=5(4)3−4(4)−1=5(64)−16−1=320−17=303\begin{aligned} g(4) &= 5(4)^3 - 4(4) - 1 \\ &= 5(64) - 16 - 1 \\ &= 320 - 17 = \mathbf{303} \end{aligned}
f(5)=5(5)2−3(5)−1=5(25)−15−1=125−16=109\begin{aligned} f(5) &= 5(5)^2 - 3(5) - 1 \\ &= 5(25) - 15 - 1 \\ &= 125 - 16 = \mathbf{109} \end{aligned}
L(−5)=−3(−5)+26−3(−5)=15+26+15=1721L(-5) = \dfrac{-3(-5)+2}{6-3(-5)} = \dfrac{15+2}{6+15} = \mathbf{\dfrac{17}{21}}
m(5)=43(5)2−3=43(25)−3=472=118m(5) = \dfrac{4}{3(5)^2-3} = \dfrac{4}{3(25)-3} = \dfrac{4}{72} = \mathbf{\dfrac{1}{18}}
b(6y3)=−4(6y3)2=−4(36y6)=−144y6b(6y^3) = -4(6y^3)^2 = -4(36y^6) = \mathbf{-144y^6}
g(3x2)=5(3x2)3−4(3x2)−1=5(27x6)−12x2−1=135x6−12x2−1\begin{aligned} g(3x^2) &= 5(3x^2)^3 - 4(3x^2) - 1 \\ &= 5(27x^6) - 12x^2 - 1 \\ &= \mathbf{135x^6 - 12x^2 - 1} \end{aligned}
b(a+h)=−4(a+h)2=−4(a2+2ah+h2)=−4a2−8ah−4h2\begin{aligned} b(a+h) &= -4(a+h)^2 = -4(a^2 + 2ah + h^2) \\ &= \mathbf{-4a^2 - 8ah - 4h^2} \end{aligned}
t(a−h)=7(a−h)−4=7a−7h−4t(a-h) = 7(a-h) - 4 = \mathbf{7a - 7h - 4}

Ejercicios extra

▼
Bloque 7

dadas las funciones

b(x)=−6x2b(x) = -6x^2
f(x)=3x2+4x−2f(x) = 3x^2 + 4x - 2
t(x)=4x−3t(x) = 4x - 3
m(x)=32x2−5m(x) = \dfrac{3}{2x^2-5}
g(x)=2x3−5x+4g(x) = 2x^3 - 5x + 4
L(x)=2x+57−4xL(x) = \dfrac{2x+5}{7-4x}

encuentra

g(−3)g(-3) f(4)f(4) L(−2)L(-2) m(−3)m(-3) b(3y5)b(3y^5) g(−2x2)g(-2x^2) b(a+h)b(a+h) t(a−h)t(a-h)
Bloque 8

dadas las funciones

b(x)=7x2b(x) = 7x^2
f(x)=−2x2+6x+1f(x) = -2x^2 + 6x + 1
t(x)=−5x+6t(x) = -5x + 6
m(x)=64x2+3m(x) = \dfrac{6}{4x^2+3}
g(x)=4x3−3x+2g(x) = 4x^3 - 3x + 2
L(x)=x−62+5xL(x) = \dfrac{x-6}{2+5x}

encuentra

g(−1)g(-1) f(−4)f(-4) L(4)L(4) m(2)m(2) b(−4y3)b(-4y^3) g(5x2)g(5x^2) b(a+h)b(a+h) t(a−h)t(a-h)
Bloque 9

dadas las funciones

b(x)=−5x2b(x) = -5x^2
f(x)=6x2−2x+3f(x) = 6x^2 - 2x + 3
t(x)=8x−5t(x) = 8x - 5
m(x)=54x2−9m(x) = \dfrac{5}{4x^2-9}
g(x)=−4x3+x−3g(x) = -4x^3 + x - 3
L(x)=4x+32−3xL(x) = \dfrac{4x+3}{2-3x}

encuentra

g(2)g(2) f(−2)f(-2) L(−1)L(-1) m(4)m(4) b(2y4)b(2y^4) g(−x2)g(-x^2) b(a+h)b(a+h) t(a−h)t(a-h)
Bloque 10

dadas las funciones

b(x)=3x2b(x) = 3x^2
f(x)=−x2+5x−4f(x) = -x^2 + 5x - 4
t(x)=−2x+7t(x) = -2x + 7
m(x)=75+3x2m(x) = \dfrac{7}{5+3x^2}
g(x)=6x3+2x−5g(x) = 6x^3 + 2x - 5
L(x)=−x+43+2xL(x) = \dfrac{-x+4}{3+2x}

encuentra

g(−2)g(-2) f(6)f(6) L(2)L(2) m(−2)m(-2) b(−3y3)b(-3y^3) g(4x2)g(4x^2) b(a+h)b(a+h) t(a−h)t(a-h)
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