Knowing the function fx x lnx, find the equation for the tangent l l of the function fx at point 1 1

Its derivative is f'(x)=1/x-a/x²

When a 0, f'(x) 0, monotonically increasing, no extremum.

When a 0, when x=a, it is the extreme point, f(a)=lna+1 look, thank you!!

1) f(x)=x-lnx

f'(x)=1-1/x

Tangent slope at (1,1):

f'(1)=1-1=0

Then the equation for the tangent l: y-1=f'(1) (x-1) gives the equation for the tangent l: y=1

2) Order f'(x)=0

Get 1-1 x=0

x=1f(x) is defined in the domain (0,+

When 01, f'(x)>0 monotonically increased.

The derivative of fx is 1-1 x, and the derivative at (1,1) is 0, so the tangent is y=1

As for the monotonic interval, let the derivative function 》0, which is the increasing interval, i.e., x>1, and the decreasing interval 0

k=1-1=0

l:y=if x belongs to (1, positive infinity) df(x) dx>0, i.e. f(x) increases monotonically.

If x belongs to (0,1) d(x) dx<0

That is, f(x) is monotonically decreasing.

Tangent equation y=1

Negative infinity, 0) monotonically increases, (0, 1) monotonically decreases, (1, positive infinity) monotonically increases.

Answer: f(x)=lnx+x 2

Derivative: f'(x)=1/x+2x

When x=1: f(1)=0+1 2=1

f'Modified eggplant macro (1) = 1 1 + 2 * 1 = 3

Therefore, the tangent point is the nuclear canona (1,1), the slope is k=3, the tangent equation is y-1=3(x-1), and y=3x-2

f'(x)=1+2 xln2lnx+(2 x) Wu do xf'(1)=1+2=3

The equation of the circumcised balance line at (1, 1) is y=3(x-1)+1, i.e., y=3x-2

y'=1/x

The tangent slope is k=y'(1)=1/1=1

The tangent equation is.

y-ln3=1·(x-1)

That is: y=x-1+ln3

f'(x)=e^x(lnx-1)+(e^x+1)*(1/x)f'(1)=e+1

f(1)=0

Tangent equation: y=(e+1)*x

If it is not to the x power of e, it is e multiplied by x

Then f'(x)=e(lnx-1)+(ex+1)*(1/x)f'(1)=e+1

f(1)=0

Tangent equation: y=(e+1)*x

y=f(x)=lnx+1/ex

f'(x)=1/x-1/ex²

Slope of the tangent at the point (1, f(1)) = f'(1)=1-1 e The tangent equation is y-f(1)=(1-1 e)(x-1), i.e., (1 e-1)x+y+1-2 e=0