c98c9033-1bb0-4f70-a1c1-cb043c71bb89 22 3 5 false false 0 0 decimal

&[DATE] &[TIME] - &[FILENAME]

f.n k Q M k f.n 2^* 1 k + (f.n 2^* 1 - (2^Q k * f.n * f.n 2^1 - (2^* + sqrt / :

Q 0.5 :

k 5 :

f.n k Q M

You need to keep identifier of the variable for differentiation separate from the identifier of the solution variable,

sln f.n k Q M f.n diff f.n f.n0 ≡ at f.n0 01 solve : 0 0.546 21 mat

sln 2 el k Q M 1.31

f.n k Q M f.n diff f.n f.n0 ≡ at f.n0 0.5 roots 0.546

# # at

Use the substitution function at() to first perform the derivative and then find it's zero.

alternatively first define the derivative as a function and then solve

M' f.n k Q M f.n diff :

sln M' f.n 01 solve : 0 0.546 21 mat

M' f.n 0.5 roots 0.546