помогите пожалуйта, очень надо

Question 1

Question 2

1)

$f(x)=100x^{10}-10x^{100}$

$f'(x)=1000x^{9}-1000x^{99}=1000(x^{9}-x^{99})$

$f'(1)=1000(1^{9}-1^{99})=0$

2)

$f(x)=x^2-3x+1$

$f'(x)=2x-3$

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</p>

$f(x)=\frac{x-3}{2x+5}$

$f'(x)=\frac{(x-3)'(2x+5)-(x-3)(2x+5)'}{(2x+5)^2}=\frac{2x+5-2(x-3)}{(2x+5)^2}=\frac{2x+5-2x+6}{(2x+5)^2}=\frac{11}{(2x+5)^2}$

f'(x)=0, $\frac{11}{(2x+5)^2}=0$ , нет решений

f'(x)>0, 0" alt="\frac{11}{(2x+5)^2}>0" align="absmiddle" class="latex-formula">, 2x+5≠0, x≠-2,5, x∈(-∞;-2,5)U(2,5;+∞)

f'(x)<0, <img src="https://tex.z-dn.net/?f=%5Cfrac%7B11%7D%7B%282x%2B5%29%5E2%7D%3C0" id="TexFormula10" title="\frac{11}{(2x+5)^2}<0" alt="\frac{11}{(2x+5)^2}<0" align="absmiddle" class="latex-formula">, нет решений

3)

$f(x)=3x-4x^3$

$f'(x)=3-12x^2$

$f'(1)=3-12\cdot1^2=-9$

$f'(5)=3-12\cdot5^2=3-12\cdot25=-472$

$f'(x+2)=3-12(x+2)^2=3-12(x^2+4x+4)=3-12x^2-48x-48=-12x^2-48x-45$

4)

$f(x)=x^5-2\sqrt{x}$

$f'(x)=5x^4-2\frac{1}{2\sqrt{x}}=5x^4-\frac{1}{\sqrt{x}}$

$f(x)=\frac{x^2-1}{x^2+1}$

$f'(x)=\frac{(x^2-1)'(x^2+1)-(x^2-1)(x^2+1)'}{(x^2+1)^2}=\frac{2x(x^2+1)-2x(x^2-1)}{(x^2+1)^2}=\frac{2x(x^2+1-x^2+1)}{(x^2+1)^2}=\frac{4x}{(x^2+1)^2}$

arsenlevadniy_zn · Answer 1 · 2018-02-06T04:13:48+0000

1)