Question

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

Answer 1

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

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

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

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

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

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