设二次函数f=mx^2+nx+t的图线过原点,g=ax^3+bx-3(x>0) f,g的导函数为f'和g',且f'=0f‘=-2 f=g,f'=g'是否存在实常数k和m,使得f(x)≥kx+m和g(x)≤kx+m?若存在,求k和m的值,若不存在,说明理由
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![设二次函数f=mx^2+nx+t的图线过原点,g=ax^3+bx-3(x>0) f,g的导函数为f'和g',且f'=0f‘=-2 f=g,f'=g'是否存在实常数k和m,使得f(x)≥kx+m和g(x)≤kx+m?若存在,求k和m的值,若不存在,说明理由](/uploads/image/z/9297607-31-7.jpg?t=%E8%AE%BE%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%3Dmx%5E2%2Bnx%2Bt%E7%9A%84%E5%9B%BE%E7%BA%BF%E8%BF%87%E5%8E%9F%E7%82%B9%2Cg%3Dax%5E3%2Bbx-3%28x%3E0%29+f%2Cg%E7%9A%84%E5%AF%BC%E5%87%BD%E6%95%B0%E4%B8%BAf%27%E5%92%8Cg%27%2C%E4%B8%94f%27%3D0f%E2%80%98%3D-2+f%3Dg%2Cf%27%3Dg%27%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E5%AE%9E%E5%B8%B8%E6%95%B0k%E5%92%8Cm%2C%E4%BD%BF%E5%BE%97f%28x%29%E2%89%A5kx%2Bm%E5%92%8Cg%28x%29%E2%89%A4kx%2Bm%3F%E8%8B%A5%E5%AD%98%E5%9C%A8%2C%E6%B1%82k%E5%92%8Cm%E7%9A%84%E5%80%BC%2C%E8%8B%A5%E4%B8%8D%E5%AD%98%E5%9C%A8%2C%E8%AF%B4%E6%98%8E%E7%90%86%E7%94%B1)
设二次函数f=mx^2+nx+t的图线过原点,g=ax^3+bx-3(x>0) f,g的导函数为f'和g',且f'=0f‘=-2 f=g,f'=g'是否存在实常数k和m,使得f(x)≥kx+m和g(x)≤kx+m?若存在,求k和m的值,若不存在,说明理由
设二次函数f=
mx^2+nx+t的图线过原点,g=ax^3+bx-3(x>0) f,g
的导函数为f'和g',且f'=0
f‘=-2 f=g,f'=g'是否存在实常数k和m,使得f(x)≥kx+m和g(x)≤kx+m?若存在,求k和m的值,若不存在,说明理由
设二次函数f=mx^2+nx+t的图线过原点,g=ax^3+bx-3(x>0) f,g的导函数为f'和g',且f'=0f‘=-2 f=g,f'=g'是否存在实常数k和m,使得f(x)≥kx+m和g(x)≤kx+m?若存在,求k和m的值,若不存在,说明理由
f(x)=mx^2+nx+t,f(0)=t=0
f(x)=mx^2+nx
f'(x)=2mx+n
g(x)=ax^3+bx-3 (x>0)
g'(x)=3ax^2+b
f'(-1)=n-2m=-2 ①
f(1)=g(1)==> m+n=a+b-3 ②
f'(1)=g'(1)==> 2m+n=3a+b ③
①②③==>
n=2m-2 ,a=(m-3)/2 ,b=5(m+1)/2
f(x)=mx^2+(2m-2)x
g(x)=(m-3)/2x^3+5(m+1)/2x-3
当m=3时,
g(x)=10x-3 ,f(x)=3x^2+4x
g(x)=f(x)解得(x-1)^2=0
∴g(x)是f(x)在(1,7)点处的切线
f(x)≥10x-3恒成立,g(x)=10x-3
此时,符合题意,k=10,m=3
f'(x)=6x+4,f'(x)=10==>x=1
m≠3时,
f(x)为二次函数,若f(x)≥kx+m恒成立
f(x)图像开口必需朝上,因此需m>0
g(x)为三次函数,若g(x)≤kx+m恒成立,
需x^3系数(m-3)/2