Question

40 pts !! f(z)= (2z^3-3z+√z-1)/z
find f'(1/4)

Answer 1

f (1/4) = 13 this is the answer

Answer 2

`DERIVATIVES \\ \\ \\ We're \: given \: that \: , \\ \: \\ f(z) \: = \frac{2 {z}^(3) - 3z + √(z) - 1}{z} \\ \\ f(z) \: = 2 {z}^(2) - 3 + (1)/( √(z) ) - (1)/(z) \\ \\ f'(z)= 4z - 0 - (1)/(2z √(z) ) - ( - \frac{1}{ {z}^(2) } ) \\ \\ f'(z)= 4z - (1)/(2z √(z) ) + \frac{1}{ {z}^(2) } \\ \\ f'( (1)/(4) ) = 4 ((1)/(4) ) - \frac{1}{2( (1)/(4) )( \sqrt{ (1)/(4) )} } + \frac{1}{ {( (1)/(4) )}^(2) } \\ \\ f'( (1)/(4) ) = 1 - 4 + 16 \\ \: \\ \\ f'( (1)/(4) ) = 13 \: \: \: \: \: \: \: \: \: \: Ans.`