Question

Найдите: 1) f`(0)-f`(1), если f(x)=13x^2-7x+5; 2) f`(1)-f`(9), если f(x)=sqrt{x}+x; 3) f`(4)-f`(9), если !

Answer 1

$1) \ f(x)=13x^{2} -7x+5\\\\f'(x)=13(x^{2})'-7(x)'+5'=26x-7\\\\f'(0)=26\cdot 0-7=-7\\\\f'(1)=26\cdot 1-7=19\\\\\boxed{f'(0)-f'(1)=-7-19=-26}\\\\\\2) \ f(x)=\sqrt{x} +x\\\\f'(x)=(\sqrt{x})'+x'=\dfrac{1}{2\sqrt{x} }+1\\\\f'(1)=\dfrac{1}{2\sqrt{1} }+1=\dfrac{1}{2}+1=1\dfrac{1}{2} \\\\f'(9)=\dfrac{1}{2\sqrt{9} }+1=\dfrac{1}{6}+1=1\dfrac{1}{6}\\\\\boxed{f'(1)-f'(9)=1\dfrac{1}{2}-1\dfrac{1}{6}=\dfrac{3}{6} -\dfrac{1}{6}=\dfrac{2}{6}=\dfrac{1}{3}}$

$3) \ f'(x)=\dfrac{1}{2\sqrt{x} }+1\\\\f'(4)=\dfrac{1}{2\sqrt{4} } +1=\dfrac{1}{4}+1=1\dfrac{1}{4}\\\\f'(9)=1\dfrac{1}{6} \\\\\boxedf'(4)-f'(9)=1\dfrac{1}{4}-1\dfrac{1}{6} =\dfrac{3}{12}-\dfrac{2}{12} =\dfrac{1}{12}$

Answer 2

$1)\ \ f(x)=13x^2-7x+5\ \ ,\ \ \ f'(x)=26x-7\ \ ,\\\\f'(0)-f'(1)=(0-7)-(26-7)=\boxed{-26\ }\\\\\\2)\ \ f(x)=\sqrt{x}+x\ \ ,\ \ \ f'(x)=\dfrac{1}{2\sqrt{x}}+1\ \ ,\\\\f'(1)-f'(9)=\Big(\dfrac{1}{2} +1\Big)-\Big(\dfrac{1}{6}+1\Big)=\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{2}{6}=\boxed{\dfrac{1}{3}\ }\\\\\\3)\ \ f(x)=\sqrt{x}+x\ \ ,\ \ \ f'(x)=\dfrac{1}{2\sqrt{x}}+1\ \ ,\\\\f'(4)-f'(9)=\Big(\dfrac{1}{4} +1\Big)-\Big(\dfrac{1}{6}+1\Big)=\dfrac{1}{4}-\dfrac{1}{6}=\boxed{\dfrac{1}{12}\ }$