Find the minimum value of the function y 2 lg x 2 lg x 3

lg(x-2)^2-lg(x-3)

lg((x-2)^2/(x-3))

x-3)+2+1/(x-3)

x-2)^2>=0

LG increases at (0, +infinity)! x>3

Therefore, (x-2) 2=0 is the minimum is 4

So the minimum value of the function y=2lg(x-2)-lg(x-3) is lg4=2lg2. Thank you

y=2lg(x-2)-lg(x-3)=lg(x-2) 2 (x-3), where x>3

lg [ x-3)+1 (x-3)+2 ] basic inequality) = lg 4

The equal sign holds if and only if x-3=1 (x-3) => x=4

lg(x-2)^2/(x-3)>0

x>3

LG monotonically increased.

Just ask for (x-2) 2 (x-3).

x-2)^2/(x-3)

x-3+1)^2/(x-3)

x-3)+2+1/(x-3)

2 root number [(x-3)*1 (x-3)]+2

Therefore, the minimum value of (x-2) 2 (x-3) is 4 so. y=2lg(x-2)-lg(x-3) and the minimum value is lg4

y=lg(x-2)^2/(x-3)

lg(x^2-4x+4)/(x-3)

lg[(x-3)^2+2(x-3)-5]/(x-3)lg[(x-3)+2-5/(x-3)]

x-3)+2-5 (x-3) has a minimum value of 6

The minimum value of y is LG6

y=lg(x 3) lg(x 12)=(lgx-lg3)(lgx-lg12)=(lgx) 2-(lg3+lg12)lgx+lg3lg12=(lgx) 2-2lg6lgx+lg3lg12=(lgx-lg6) 2+lg3lg12-(lg6) 2=(lgx-lg6) 2-(lg2) 2 Therefore, when x=6, there is a great reputation for the most Shichang small value - (lg2) 2

First change the form of the divination function to change:

y=lgx 3 *lgx 12=lgx 3 *(lgx 3+lg1 4)=(lgx loose slag 3)*(lgx 3)+(lgx 3)*(lg1 4).

Let Z=LGX 3, then the above equation becomes: Y=Z*Z+Z*LG1 4 This is a unary quadratic equation whose flushing quietly a=1,b=lg1 4,c=0 Therefore, when Z=-B 2A=-(1 2)*Lg1 4=-lg1 2, Y has a minimum value.

Thus, z=lgx 3=-lg1 2

The value of x can be solved: x=6

In this case, y=lgx 3 *lgx 12=-lg2*lg2A: When x is 6, y=lgx 3 *lgx 12 has a minimum value, and the minimum value is: -lg2*lg2

y=LG(2x 2+18)-LG3x(x>0)=LG((2x 2+18) brightness(3x))=LG(2x 3+6x).

Because 2x 3+6 x "2 times the root number (Chang Lack Width 2x 3*6 x) = 2 times the root number under 4=4

and y=lgx is an incrementing function.

So the smallest value of y is y=lg4

y=lg x hail album 3*lg x 12=(lgx-lg3)*(lgx-lg12)set to be unscrupulouslgx=ty=(t-lg3)(t-lg12)=t 2-(lg3+lg12)t+lg3*lg12=t 2-(lg3+lg4+lg3)t+lg3lg12=t 2-2(lg2+lg3)t+lg3*lg12=(t-lg2-lg3) 2+lg3*(2lg2+lg3)- (lg2+lg3) 2=(t-lg6) 2-(lg2) 2t=lgx=lg6 x=6, the minimum value is -(lg2) 2

Find the derivative, then make the derivative equal to zero, and see if there is an extreme point (in which case x should be positive). If so, then judge the second derivative and see if it is a maximum or a minimum. That's it.

Lg(m n)=lgm-lgn is used for simplification.

y=lg(x/3)lg(x/12)

lgx-lg3)(lgx-lg12)

LGX) 2-(LG3+LG12)LGX+LG3LG12 treats LGX as a whole.

Let t=lgx

then y=t 2-(lg3+lg12)t+lg3lg12=t 2-(lg36)t+lg3lg12

t^2-(2lg6)t+lg3lg12

The opening is symmetrically pumped upwards as t=lg6

The minimum value is at the axis of symmetry.

Substitute t=lg6.

y(min)=-(lg2) 2 can be obtained

If you don't understand, you can ask.

The guy above me made a mistake in the fourth line.

According to ln(x y)=lnx-lny

y=lg(x/3)lg(x/12)

y=(lnx-ln3)(lnx-ln12)y=(lnx-ln3)(lnx-2ln3)y=(lnx) -3ln3lnx+2(ln3) let t=lnx, then t r

y=t²-3ln3t+2(ln3)²

Formula y=(y=(When t=, the minimum value is obtained.

At this point, t=lnx=

So x=3

Agree with the second person's answer.

That's right.

ln12=2ln3 This is a "genius job".

ln12=ln3*4=ln3+ln4=ln3+ln2^2=ln3+2ln2

Point out that others are wrong, and explain further.

y=lg(x/3)* lg(x/12)

lgx-lg3)*(lgx-lg12)=(lgx)^2-(lg3+lg12)*1gx+lg3*1g12=(lgx-1/2(lg3+lg12))^2+lg3*lg12-1/4(lg3+lg12)^2

When LGX=1 2(LG3+LG12)=1 2LG36=1 2LG6 2=1 2*2LG6=LG6

i.e. when x=6.

The minimum value of the function is:

y=lg3*lg12-1/4(lg3+lg12)^2=lg3*lg12-1/4(lg36)^2=lg3*lg12-1/4*4*(lg6)^2=lg3lg12-lg6*lg6

lg3*(lg3*2^2)-(lg2*3)^2=lg3(lg3+2lg2)-(lg3+lg2)^2=-(lg2)^2

What grade are you in now? Has derivative math ever been? Try it with a derivative.