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f(x)=(x-5)lnx. Aflati ecuatia tangentei in punctul A(1,0). Multumesc

Răspuns :

f(x)(x-5)lnx x>0
f `(x)=lnx+(x-5)/x
f `(1)=ln`1+(1-5)/1=-4
y-yA=f `(xA)(x-xA)
y-0= -4*(x-(-4))
y= -4x-16