Question

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

Answer 1

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

Answer 2

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