王俊凱 asked in 科學數學 · 6 years ago

微積分~~~~求救please~~~

1.Find the absolute maximum and absolute minimum values of f(t)=t(4-t^2)^1/2 on the interval [-1,2].

2.Prove that the function f(x)=x^101+x^51+x+1 has neither a local maximum nor a

local maximum .

3.Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2].

4.Does there exist a function f such that f(0)=-1,f(2)=4, and f ' (x)<=2 for all x ?

5. (1).Find all points on the curve x^2y^2+xy=2 where the slope of the tangnet

line is -1.

(2).If f(x)+x^2[f(x)^3]=10 and f(1)=2,find f ' (1).

6.Use a linear approximation(or differentials) to esitmate (1.999)^4.

7.Find the critical numbers of the function.

(1). g(t)=t^4+t^3+t^2+1 (2).F(x)=x^4/5(x-4)^2

8.Let f(x)=(x-3)^-2.Show that there is no value of c in (1,4) such that

f(4)-f(1)=f ' (c)(4-1). Why does this not contradict the Mean Value Theorem?

9.If f(1)=10 and f ' (x)大於等於2 for 1<=x<=4,how small can f(4) possibly be ?

10.Find the absolute maximum and absolute minimum values of

f(x)=2x^3-3x^2-12x+1 on the interval [-2,3].

1 Answer

Rating
  • 麻辣
    Lv 7
    6 years ago
    Favorite Answer

    1.f'(t)=√(4-t^2)-t*0.5*2t/√(4-t^2)=[(4-2t^2)-t^2]/√(4-2t^2)=(4-3t^2)/√(4-2t^2)=0t^2=4/3 => t=+-2/√3=+-1.155=> Extreme fall on [-1,2]

    2. f'(x)=101x^100+51x^50+1=0=> 0=101y^2+51y+1......y=x^50=> y=(-51+-√(2601-404))/202=(-51+-√2197)/202=(-51+-46.87)/202=(-97.87 -4.13)/202=(-0.4845 -0.0204)x=y^(1/50)=(-0.4845 -0.0204)^(1/50)=Imaginary=> It has without max.

    3.y'=3x^2-15=0 => x=+-√5=+-2.24y"=6x => y"(+-√5)=+-6√5 => max or min=> The most root exist [-2,2]

    4. Assume f(x)=ax^2+bx+c for having 3 conditions.f(0)=c=-1f(2)=4a+2b-1=4 => b=5/2-2af'(x)=2ax+b=2ax+5/2-2a=2a(x-1)+5/2<=2=> 2a(x-1)<=-1/2=> It is uncertain

    5 (2)'=[(xy)^2+(xy)]'0=2(xy)*(xy'+y)+(xy'+y)=2x^2*yy'+2xy^2+xy'+y=-2yx^2+2xy^2+y-x=2xy(y-x)+(y-x)=(y-x)(1-2xy)=> x=y, xy=1/2=> x^2=y^2=1/2=> x=y=+-1/√2

    11. 0=f'(x)+2x*f(x)^3+x^2*3*f(x)^2*f'(x)=[1+3x^2*f(x)^2]*f'(x)+2x*f(x)f'(x)=-2x*f(x)/[1+3x^2*f(x)^2]f'(1)=-2*1*f(1)/[1+3*1*f(1)^2]=-2*2/(1+3*4)=-4/13

    6. 1.999^4=(2-0.001)^4=2^4-4*2^3*0.001+6*2^2*0.001^2-4*2*0.001^3+0.001^4=16-32/1000+24/10^6-8/10^9+1/10^12≒16-0.032=15.968

    7.1. g(t)=t^4+t^3+t^2+1 g'(t)=4t^3+3t^2+2t=t(4t^2+3t+2)=0t1,t2=[-3+-√(9-32)]/8=(-3+-√23*j)/8t3=0......ans

    7.2 F(x)=x^4/5(x-4)^2 x^4=F*5(x-4)^24x^3=F'*5(x-4)^2+5F*2(x-4)F'=[4x^3-10(x-4)F]/5(x-4)^2=0=> F=2x^3/5(x-4)=x^4/5(x-4)^2=> 2=x/(x-4)=> x=8

    8. Because lim(x->3+)f(x)=+∞, lim(x->3-)f(x)=-∞It is not continuous.

    9. Assume f(x)=ax+b for two conditions.f'(x)=a>=2 => a=2+k; k>=0f(1)=a+b=10 => b=10-a=8-kf(x)=(k+2)x+(8-k)f(4)=4(k+2)+8-k=3k+16Min.f(4)=16......k=0

    10. f'(x)=6x^2-6x-12=6(x^2-x-2)=6(x+1)(x-2)=0x=-1, 2f"(x)=12x-6=6(2x-1)f"(-1)=-18<0 => maxf"(2)=18>0 => minmax=f(-1)=-2-3+12+1=8min=f(2)=16-12-24+1=-13

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