Вопрос пользователя:
помогите пожалуйта, очень надо
Илюха отвечает:
1)
=100x^{10}-10x^{100} "f(x)=100x^{10}-10x^{100}")
=1000x^{9}-1000x^{99}=1000()
x^{9}-x^{99})” title=”f'(x)=1000x^{9}-1000x^{99}=1000(x^{9}-x^{99})” alt=”f'(x)=1000x^{9}-1000x^{99}=1000(x^{9}-x^{99})” />
=1000()
1^{9}-1^{99})=0” title=”f'(1)=1000(1^{9}-1^{99})=0” alt=”f'(1)=1000(1^{9}-1^{99})=0” />
2)
=x^2-3x+1 "f(x)=x^2-3x+1")
=2x-3 "f'(x)=2x-3")
f'(x)=0, 2x-3=0, x=1,5
f'(x)>0, 2x-3>0, x>1,5
f'(x)=0, 2x-3<0, x<1,5
=frac{x-3}{2x+5} "f(x)=frac{x-3}{2x+5}")
=frac{(x-3)'(2x+5)-(x-3)(2x+5)'}{(2x+5)^2}=frac{2x+5-2(x-3)}{(2x+5)^2}=frac{2x+5-2x+6}{(2x+5)^2}=frac{11}{(2x+5)^2} "f'(x)=frac{(x-3)'(2x+5)-(x-3)(2x+5)'}{(2x+5)^2}=frac{2x+5-2(x-3)}{(2x+5)^2}=frac{2x+5-2x+6}{(2x+5)^2}=frac{11}{(2x+5)^2}")
f'(x)=0, ^2}=0 "frac{11}{(2x+5)^2}=0")
, нет решений
f'(x)>0, ^2}<0 "frac{11}{(2x+5)^2}<0")
, нет решений
3)
=3x-4x^3 "f(x)=3x-4x^3")
=3-12x^2 "f'(x)=3-12x^2")
=3-12cdot1^2=-9 "f'(1)=3-12cdot1^2=-9")
=3-12cdot5^2=3-12cdot25=-472 "f'(5)=3-12cdot5^2=3-12cdot25=-472")
=3-12(x+2)^2=)
3-12(x^2+4x+4)=3-12x^2-48x-48=-12x^2-48x-45” title=”f'(x+2)=3-12(x+2)^2=3-12(x^2+4x+4)=3-12x^2-48x-48=-12x^2-48x-45” alt=”f'(x+2)=3-12(x+2)^2=3-12(x^2+4x+4)=3-12x^2-48x-48=-12x^2-48x-45” />
4)
=x^5-2sqrt{x} "f(x)=x^5-2sqrt{x}")
=5x^4-2frac{1}{2sqrt{x}}=)
5x^4-frac{1}{sqrt{x}}” title=”f'(x)=5x^4-2frac{1}{2sqrt{x}}=5x^4-frac{1}{sqrt{x}}” alt=”f'(x)=5x^4-2frac{1}{2sqrt{x}}=5x^4-frac{1}{sqrt{x}}” />
=frac{x^2-1}{x^2+1} "f(x)=frac{x^2-1}{x^2+1}")
=frac{(x^2-1)'(x^2+1)-(x^2-1)(x^2+1)'}{(x^2+1)^2}=frac{2x(x^2+1)-2x(x^2-1)}{(x^2+1)^2}=frac{2x(x^2+1-x^2+1)}{(x^2+1)^2}=frac{4x}{(x^2+1)^2} "f'(x)=frac{(x^2-1)'(x^2+1)-(x^2-1)(x^2+1)'}{(x^2+1)^2}=frac{2x(x^2+1)-2x(x^2-1)}{(x^2+1)^2}=frac{2x(x^2+1-x^2+1)}{(x^2+1)^2}=frac{4x}{(x^2+1)^2}")