Question

Помогииитеее оооочень нужнооооооооо

Answer 1

1) $f(x)= frac{x^3-3x}{1+4x^5} \ f'(x)= frac{(x^3-3x)'(1+4x^5)-(x^3-3x)(1+4x^5)'}{(1+4x^5)^2}= \ frac{(3x^2-3)(1+4x^5)-(x^3-3x)*20x^4}{(1+4x^5)^2}=$
$=frac{3x^2-3+12x^7-12x^5-20x^7+60x^5}{(1+4x^5)^2}=frac{-8x^7+48x^5+3x^2-3}{(1+4x^5)^2}$

2) $frac{1+tg alpha }{1+ctg alpha }+tg alpha = frac{1+ frac{sin alpha }{cos alpha } }{1+ frac{cos alpha }{sin alpha } }+tg alpha=frac{ frac{cos alpha +sin alpha }{cos alpha } }{ frac{sin alpha +cos alpha }{sin alpha } }+tg alpha= \ = frac{sin alpha (cos alpha +sin alpha )}{cos alpha (sin alpha +cos alpha )} +tg alpha= frac{sin alpha }{cos alpha } +tg alpha=tg alpha+tg alpha=2tg alpha$

3) $f(x)= frac{1}{3}x^3-4x^2+7x+18$


f'(x)=x²-8x+7
f'(x)=x²-8x+7=0
D=8²-4*7=64-28=36
√D=6
x₁=(8-6)/2=1
x₂=(8+6)/2=7

f'(x)=x²-8x+7>0
x²-8x+7=(x-1)(x-7)>0
x∈(-∞;1)∪(7;+∞)