Help solve a calculus problem, solve a calculus problem, thank you

Answer: Let f(t)=t(1-2t)(1-3t) t [0,1].

It is advisable to let f(t)=t(1-2t)(1-3t) a(3t-1) be established in [0,1] constantly, and determine a first

Because the inequality sought is equal at x=y=z=1 3.

Therefore, f(t)=t(1-2t)(1-3t)-a(3t-1) takes the minimum value when t=1 3, and the derivative is 0

Therefore 18t 2-10t+1-3a=0 has a root of x1=1 3, so x2=2 9, a=25 81

So g(t) increases monotonically at [0,2 9], [1 3,1] and decreases monotonically at [2 9,1 3].

So the minimum value of g(t) on [0,1] is min=0

So x(1-2x)(1-3x)+y(1-2y)(1-3y)+z(1-2z)(1-3z).

f(x)+f(y)+f(z)

25(3x-1)/81+25(3y-1)/81+25(3z-1)/81=0

Take etc if and only if x=y=z=1 3.

f(x)=2x^3-9ax^2+12a^2x

a=1, then there is f(x)=2x 3-9x 2+12x, f'(x)=6x^2-18x+12=6(x^2-3x+2)=6(x-1)*(x-2)

The tangent equation for the origin is y=kxIf the tangent coordinates are (xo,yo), then there is k=yo xo=6(xo 2-3xo+2).

yo=6(xo^3-3xo^2+2xo)=f(xo)=2xo^3-9xo^2+12xo

The solution yields 4xo 3-9xo 2=0

xo^2(4xo-9)=0

xo=0(rounded), xo=9 4

yo=2*9^3/64-9*9^2/16+12*9/4=27-729/32=135/32

Therefore, the tangent coordinates are (9 4,135 32).

So the tangent equation is y=135 72 x

x)=6x^2-18ax+12a^2=6(x-a)(x-2a)=0

Get x1=a, x2=2a

a>0, then there is f'(x) >0, function increases, in a< x<2a, f'(x) <0, function minus.

This calculus problem is to test the mastery of the basic knowledge of calculus, here d is followed by y, this is the questioner dug a small hole for the problematic, the more you can't go over, the more you can go in the opposite direction.

Derivatives on both sides of the equation with respect to y vs. z yields 3x 2y 2+2zz'(y)=z'(y) e z, so z'(y)=3x^2y^2/(e^z-2z).

Upstairs is wrong, g(x)=x * 0,x)cost 3 dt so there is:

g'(x) = (0,x)cost 3 dt+x*cosx 3, but = (0,x)cost 3 dt cannot be expressed by elementary functions.

Using the formula x 3 + y 3 = (x + y) (x 2-xy + y 2), it is equivalent to x+y in the formula that you said "x plus 3 times the root number under 1 minus x 3" is equivalent to x+y in the formula, which is essentially to rationalize the molecule, after this step is completed, it is found that the denominator is close to infinity when x is close to plus or minus infinity, so its reciprocal is close to 0, in summary, Zheng San, the limit is zero.