Question

РЕБЯТ ПЖ СРОЧНО!ПОМОГИТЕ С 87-90.БУДУ ОЧЕНЬ БЛАГОДАРНА

Answer 1

87
$f(x)=\sqrt{2x+1}\\\\ f(x_0)=f(7.5)=\sqrt{2*7.5+1}=\sqrt{15+1}=\sqrt{16}=\sqrt{4^2}=4$

$f'(x)=[\sqrt{2x+1}]'=\frac{(2x+1)'}{2\sqrt{2x+1}}=\frac{1}{\sqrt{2x+1}}\\\\ f'(7.5)=\frac{1}{\sqrt{15+1}}=\frac{1}{4}=0.25$

88
$(x)=\sqrt{5x+1}\\\\ f(x_0)=f(3)=\sqrt{5*3+1}=\sqrt{15+1}=\sqrt{16}=\sqrt{4^2}=4$

$f'(x)=[\sqrt{5x+1}]'=\frac{(5x+1)'}{2\sqrt{5x+1}}=\frac{5}{2\sqrt{5x+1}}\\\\ f'(3)=\frac{5}{2*4}=\frac{5}{8}=0.625$

89
$y(x)=\frac{x}{x^2+1}\\\\ y(x_0)=y(0)=\frac{0}{0^2+1}=\frac{0}{1}=0$

$y'(x)=[\frac{x}{x^2+1}]'=\frac{1*(x^2+1)-x*2x}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}\\\\ y'(0)=\frac{1-0}{1}=1$

90
$y(x)=\frac{4x-7}{x^2+4}\\\\ y(x_0)=y(0)=\frac{4*0-7}{0^2+4}=\frac{-7}{4}=-\frac{7}{4}=-1.75$

$y'(x)=[\frac{4x-7}{x^2+4}]'=\frac{4*(x^2+4)-(4x-7)*2x}{(x^2+4)^2}=\frac{4x^2+16-8x^2+14x}{(x^2+4)^2}=\frac{-4x^2+16+14x}{(x^2+4)^2}\\\\ y'(0)=\frac{0+16+0}{(0+4)^2}=1$