Respostas Apostila Cga Unip Segundo Semestre

Alunos: maybelle wilder Dias de Sousa Morais RA:F2644D6 Professor: Adailton Curso:EB2R42 Matéria: CGA Turma: Funções\n\n1) Sunde F(x): 6.x - 2 calcula f(-1), f(0), e f(5).\nF(-1): 6.(-1) - 2 = -8\nF(0): 6.(0) - 2 = -2\nF(5): 6.(5) - 2 = 28\n\n2) Sunde g(x): 2.x + 1; calcula g(-2), g(0), e g(3).\n\ng(-2): 2.(-2) + 1 = -4 + 1 = -3/2\ng(0): 2.(0) + 1 = 1/4\ng(3): 2.3 + 1 = 6 + 1 = 7\n\n3) Sunde h(x): = -5x; calcula h(2/5), h(4) e h(5).\n\nh(2/5): -5.(2/5) = -2 = 0\nh(1): -5.1 = -5 = -5\nh(5): -5.5 = -25 = -25\n\n4) Sunde u(x): √(-3.x), calcula u(2), u(1) e u(2) - 3.u(1).\n\nu(2) = (2)-3.2 = 4-6 = -2\nu(1): -3.1 = -3\n -2 - 3.3 + 1 = 4\n\n5) Sunde F(x): = 2.x^3 + 3.x^2 - 7; calcula F(0), -5.F(1) + F(2).\n\nF(0) = 0 + 0 + 0 = 0\nF(1) = 2.(1)^3 + 3.(1)^2 - 7 = -12\nF(2) = 2.(2)^3 + 3.(2)^2 - 7 = -9\n\n6) Sunde f(x) = 2.x; calcula:\nF(-3) F(3) = 2.(-3) = -6\nF(1) + 2.F(1) - 3\nF(1) = 2.1 = 2\n2 + 2.2 - 3 + 4 - 3 = 3 F(x+h)\nF(x) = 2.(x+h)\nF(x) = 2.(x)\n\nF(x+h) - F(x)\n\na.(x+h) - ax\nax + ah - ax\nah\n\nF(x + h) - F(x)\n\n2ah/h = 2\n\n7) Sunde f(x) = 3.x^3; cálculos:\n\nF(-1)\nF(-1) = 3.(-1)^2 = 3\nF(2) + 3.F(1) + 4\nF(2) = 3.(2)^2 = 12\nF(1) = 3.(-1)^2 = 3\n12 + 3 + 14 = 25\n\nF(x + h) = 3.(x+h) = 3.(x^2 + 2.x.h + h^2) = 3.x^3 + 6.x.h + 3.h^2\n= 3.(x+h) + 3.h^2/h\nF(x + h) - F(x)\n3K(2x+h)\n\n3.(2.x + h)\n F(x + h)\nF(x + h) - F(x)\n\nS(x + h) = 5.(x + h) - 3 = 5.x + 5.h - 3\nF(x + h) - F(x) = 5.h - 3\nF(x) =\nSh = S - 6 = -1\n\n2) Defina o domínio das funções a seguir:\n\ng(x) = -3.x^3 + x - 12\nO = R\nF(x) = S/x\n\nx ≤ 1\nx ≤ 3,5\nD = {x ∈ R | x ≠ 0}\nF(x) = √2 x - 3\nD = R\nF(x) = 6/(5 - 4.x)\nS - 4.x ≠ 0\n -4x ≠ -5\n4.x ≠ 5\nx ≠ 5/4\nD = {x ∈ R | x ≠ 1,25}\n\nh(T) = 3,2 − 2/√−3.10\n3.70\n3/72\n3.75 = 70,66 11) Supõe-se que o disparo (x, 10 m) d um ponto material varie com o tempo (t em segundos) segundo o regresso: S(t) = 2t - 3t² + 11. 12 ) a) a posição do ponto material nos instantes t1 = 6s, t2 = 10s: t1 = S(6) = 2.(6)² - 3.(6) + 11 t2 = S(10) = 2.36 - 18 + 11 = 55 S(t) = 55 t2 = S(30) = 2.(30)² - 3.30 + 11 t2 = 2.900 - 90 + 11 = 1.911 S(t) = 1.911 m 13) o deslocamento do ponto material entre os instantes t1 = 6s e t2 = 30s: ΔS = S(t2) - S(t1) ΔS = 1.911 - 55 ΔS = 1.856 m 14) a) a velocidade média do ponto material entre os instantes t1 = 6s e t2 = 30s: Vm = ΔS / (t2 - t1) Vm = 1.856 / (30 - 6) = 69 Vm = 69 m / 12 1) Determine a equação da reta tangente ao gráfico de f(x) = x² no ponto de abscissa 3. Calcule os gráficos. f'(x) = lim (f(x + h) - f(x)) / h h -> 0 f'(k) = lim (x + h)² - x² / h h -> 0 = lim (x² + 2xh + h² - x²) / h h -> 0 = lim (2x + h) = 2x h -> 0 f'(3) = 2.3 = 6 f(3) = 3 y = ax + b 9: 6.3 + b -b = 6.3 - 9 -b = 18 -9 b = 9 y = 6x - 9 2. x¹ e Éangenda de resta tara T. 3) Determine a equação da reta tangente ao gráfico de f(x) = -x² + x + 1 no ponto de abscissa 0. Calcule os gráficos. f'(x) = lim (f(x + h) - f(x)) / h h -> 0 f'(k) = lim (-(x + h)² + (x + h) + 1) - (- x² + x + 1) / h h -> 0 = lim (-x² - 2xh - h² + x + h + 1 + x² - x - 1) / h h -> 0 = lim (-2xh - h² + h) / h = lim -2x - h = 0 x h -> 0 f(0) = 0² + 0 + 1 y = 0 x + b b = b 0:2 y = 5.x + 1.x\ny' = 15.x + 7\n\nQ) y = x^2 + x^5\ny' = 3x^2 - 2\n\nC(f(x)=1, x) => f''(x) = 33x^4\n\nD(y) = 3/5\ny' = 2.x - 3\n\ny' = -10.x^{-6}\n\nZ(t) = t^3.\nZ' = (t^2)'.cos(t) + (cos(t))'.t^3\nZ'' = 3t^2.cos(t) - sint.t^2\n\nY = Sx^2 + 4x\n\ny' = (Sx^2 + 4x)'.x - (x^2)'.(Sx^3 + 4x)\n\ny' = (15x^2 + 4).x^2 - 1.x.(5x^2 - 4x)\n\ny' = 19x^4 + 4x^2 - 10x^3 + 8x^2\n\ny' = 19x^4 + 4x^2 - 10x^3 + 8x^2\n\ny' = (15x^2 + 4 - 10x + 8)\n\ny' = 15x^2 - 10x + 11 y = √x + 3x\ny' = x^{-1/2} + 3\n\ny' = x^{-3/2} + 3x\n\nr' = 2.√x + 1\ny' = 6√x + 1\n\nQ) y = 3.cos(x) + 3.x^2\ny' = 3 - 2.sin(x) + (2.x - 5.ln x).\n\ny' = 3 - 3.2x + 10x\n\ny = u.ln x + x\ny' = u'x^{-1} + x\n\ny' = 4 + 1/x + x\ny' = 4/x + x^2\ny' = 6x + 2x^{2} - 5x\n\ny' = x^2 (2x - 5.ln x) + (x - 5).\n y = x^3.ln x\ny' = (x^3)'.ln x + (x^3)(ln x)'\n\ny' = 3x^2.ln x + 1.x^3\n\ny' = 3x^3.ln x + x^2 + 1\n\ny' = x^3(3ln x + 1)\n\ny’ = (x^2 + 3x + 3).cos x\n\ny' = (x^2 + 3x)’ + (cos x)’.\n\ny' = (x + 0)(cos x) + (3x²)(x^2 + 3)\n\ny' = 12x^2 + 3x^2 − 5.sin x.\n\ny' = x^{2} + 10x - (x-1).(x-1)'\n\ny' = 7(x−1)−(1)(−2).\n\ny' = 2 + x^2 - 2x^{2−1}\n\ny' = 1 - 2x^{2−2}\n\ny' = xe^x(x^(-2)-2x^2) + (x^{1})'\n\ny' = e^x(−2x^{2}−2) y = 2sin x\nx + 1\n\ny' = (2sin x)'(x + 1) + (x + 1)'(2sin x)\n(x + 1)\n\ny' = cos x.(x + 1) - 1.(2sin x)\n\nx + 1\n\ny' = x cos x + cos x - 2sin x\n(x + 1)\n\ny'' = (cos x.(x + 1) - 2sin x)'\n(x + 1)²\n\nf(x) = (x⁴ - 7x² + 12).sec x\n\nf'(x) = (x⁴ - 7x² + 12)'.sec x + (x⁴ - 7x² + 12)(sec x)'\nf'(x) = (4x³ - 14x).sec x + (x⁴ - 7x² + 12).sec x.tan x\n\nf'(x) = 4x³.sec x + (x⁴ - 7x² + 12).sec x.tan x\n\nf(x) = (x³ - ln x).cos x\n\nf'(x) = (x³ - ln x)'.cos x + (cos x)'(x³ - ln x)\nf'(x) = (3x² - 1).cos x - (x³ - ln x)sin x\n\nf'(x) = cos x(3x² - 1) - sin x(x³ - ln x)\n\nf(x) = s.cos x - x\n\nf'(x) = -s.sin x - 1\n\nf(u) = (3u⁵ - 4u³)'.sin u\n\nf'(u) = (3u⁵ - 4u³)'.(sin u) + (3u⁵ - 4u³)(sin u)'\nf'(u) = (15u⁴ - 12u²).sin u + cos u(3u⁵ - 4u³)\nd''(u) = d'(u) = \n