Suppose that
f ( x ) \displaystyle {f{{\left({x}\right)}}} f ( x ) and
g ( x ) \displaystyle {g{{\left({x}\right)}}} g ( x ) are two functions and we know that:
f ( 1 ) = − 1 \displaystyle {f{{\left({1}\right)}}}=-{1} f ( 1 ) = − 1 g ( 1 ) = 5 \displaystyle {g{{\left({1}\right)}}}={5} g ( 1 ) = 5 f ′ ( 1 ) = − 1 \displaystyle {f}'{\left({1}\right)}=-{1} f ′ ( 1 ) = − 1 g ′ ( 1 ) = 4 \displaystyle {g}'{\left({1}\right)}={4} g ′ ( 1 ) = 4
Find the following:
( f − g ) ′ ( 1 ) = \displaystyle {\left({f}-{g}\right)}'{\left({1}\right)}= ( f − g ) ′ ( 1 ) = Preview Question 6 Part 1 of 5 ( g − f ) ′ ( 1 ) = \displaystyle {\left({g}-{f}\right)}'{\left({1}\right)}= ( g − f ) ′ ( 1 ) = Preview Question 6 Part 2 of 5 ( f g ) ′ ( 1 ) = \displaystyle {\left({f}{g}\right)}'{\left({1}\right)}= ( f g ) ′ ( 1 ) = Preview Question 6 Part 3 of 5 ( f g ) ′ ( 1 ) = \displaystyle {\left(\frac{{f}}{{g}}\right)}'{\left({1}\right)}= ( g f ) ′ ( 1 ) = Preview Question 6 Part 4 of 5
If
k ( x ) = f ( x ) x 2 \displaystyle {k}{\left({x}\right)}=\frac{{{f{{\left({x}\right)}}}}}{{{x}^{{2}}}} k ( x ) = x 2 f ( x ) then
k ′ ( 1 ) = \displaystyle {k}'{\left({1}\right)}= k ′ ( 1 ) = Preview Question 6 Part 5 of 5 Submit Try a similar question
[more..]
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)