导数五步法画函数图像10个函数示意图应用举例之一
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1.函数y=(12x2+9)(4x2+14)的图像示意图：介绍函数的定义域、单调性、凸凹性、极限等性质及五点图表，并通过导数知识计算函数的单调和凸凹区间，简要画出示意图。
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2.函数y=(19x2+5)√(4x2+9)的主要性质及其图像：介绍函数的定义域、单调性、凸凹性、极限等性质，列举函数的五点图表，进一步画出函数的示意图。
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3.函数y=4√(x+80)^7图像画法及步骤：本文通过函数的定义、单调、凸凹性和极限等性质，介绍函数的主要性质及图像画法步骤。
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4.曲线x³+y³=2的主要性质及其图像示意图：介绍曲线方程的定义域、单调性、凸凹性等性质，同时用导数的知识求解函数的单调区间和凸凹区间，并简洁画出函数的图像示意图。
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5.√(x+4)+√(3y+5)=2的图像示意图：介绍曲线方程的定义域、单调性、凸凹性及极限等性质，同时用导数简洁画出函数的图像示意图。
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6.函数y=16x3+8x的图像示意图及主要性质：介绍函数的定义域、单调性、凸凹性、极限等性质，列举函数的五点图表，进一步画出函数的示意图。
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7.函数y=√(20x-87)^5图像画法及步骤：通过函数的定义、单调、凸凹和极限等性质， 并通过导数知识，介绍函数的主要性质及图像示意图画法步骤。
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8.函数y=log2(-2x+3)的图像示意图：介绍函数的定义域、单调性、凸凹性、极限等性质，列举函数的五点图表，简要画出函数的示意图。
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9.函数y=e^x(3x+4)的图像示意图：本文通过函数的定义、单调、凸凹性和极限等性质，介绍函数的主要性质及图像画法步骤。
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10.函数y=2^4x的图像示意图：介绍函数的定义域、单调性、凸凹性、极限等性质，列举函数的五点图表，进一步画出函数的示意图。
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莫戴咐4188y=根号(x的平方+x+3)的定义域值域呢 -
狄冠油15382747298 ______[答案] y=根号(x的平方+x+3)的定义域 要满足的条件是x²+x+3≥0 也即 (x+1/2)²+11/4≥0 可见 x能去任意值 所以 定义域为(-∞,+∞)
莫戴咐4188求函数y=|x - 2/x+3|的定义域、值域、单调性 -
狄冠油15382747298 ______ 函数y=|x-2/x+3|的 定义域:x≠-3、 值域:y>=0且y≠1 单调性: x-2/x+3=0 x=2 1.xx-2/x+3>0 函数递增; 2. -3x-2/x+3|x-2/x+3|=2-x/x+3 递减; 3. x>=2 x-2/x+3>0 所以 |x-2/x+3|=x-2/x+3 单调递增; 从而 单调增区间:(-∞,-3),(2,+∞) 单调减区间:(-3,2)
莫戴咐4188y=arcsin(x2 - x+3/4)的定义域 值域 -
狄冠油15382747298 ______ 因为f(x)=arcsinx的定义域是[-1,1] 令-1≤x^2-x+3/4≤1得 (1-√2)/2≤x≤(1+√2)/2 故y=arcsin(x^2-x+3/4)的定义域是[(1-√2)/2,(1+√2)/2] 又x∈[(1-√2)/2,(1+√2)/2]时 x^2-x+3/4∈[1/2,1] 所以y=arcsin(x^2-x+3/4)∈[π/6,π/2] 即值域是[π/6,π/2]
莫戴咐4188求函数y=(2x+3)/(x+1)的定义域和值域 -
狄冠油15382747298 ______[答案] 定义域:x+1≠0 x≠-1 值域:(2x+2+1)/(x+1)=2+1/(x+1) 当x>-1 (2,+∞) 当-3/2≤x<-1 [0,2) 当x>-3/2 (-∞,0)
莫戴咐4188y=(x+3)/(x - 2)的定义域 -
狄冠油15382747298 ______ 分母x-2≠0 x≠2 所以定义域是(-∞,2)∪(2,+∞)
莫戴咐4188y=x的平方+1分之x - 3 定义域 -
狄冠油15382747298 ______ 解答:y=x²+1/(x-3) 定义域只要x-3≠0即使x≠3 即使定义域为{x≠3} 点评:估计是楼主写错了,应该是x-3分之1 如果没写错,那么答案就是R也就是全体实数 如果有写错那么就是以上的答案{x≠3}
莫戴咐4188y=( - 2x+3)/(x - 2)求值域和定义域 -
狄冠油15382747298 ______[答案] y=(-2x+3)/(x-2)定义域(-∞,2)U(2,+∞) y(x-2)=-2x+3 xy-2y=-2x+3 xy+2x=2y+3 x=(2y+3)/(2+y) 值域(-∞,-2)U(-2,+∞)
莫戴咐4188函数y=3x+2/2x+3的定义域是多少?急用! -
狄冠油15382747298 ______[答案] 函数y=(3x+2)/(2x+3)的定义域是分母2x+3≠0,即x≠-3/2,也就是x≠-1.5
莫戴咐4188已知f(x)的定义域为〔0,1〕,求函数y=f(x^2)+f(x+4/3)的定义域. -
狄冠油15382747298 ______[答案] f(x)的定义域为〔0,1〕,f(x^2)的定义域为〔-1,1〕, f(x+4/3)的定义域为〔-4/3,-1/3〕,y=f(x^2)+f(x+4/3)的定义域为〔-1,-1/3〕
莫戴咐4188求下列函数的定义域:(1)y=x+8+3−x;(2)y=x2−1+1−x2x−1. -
狄冠油15382747298 ______[答案] (1)由二次根式的意义可知: x+8≥03−x≥0,解得−8≤x≤3, ∴定义域为[-8,3]. (2)由二次根式和分式的意义可知: x2−1≥01−x2≥0x−1≠0,解得x2=1且x≠1,即x=−1 ∴定义域为{-1}. 故答案为:(1)定义域为[-8,3],(2)定义域为{-1}.