Question

( y ) is Max. whan ( t=left(4 x^{2}+2 x+1right) ) is min.
( frac{d t}{d x}=0 ) (Minimumi)
( 8 x+2=0 quad-lambda x=-1 / 4 )
( begin{aligned} A t x=-1 / 4, t &=4left(frac{-1}{4}right)^{2}+2left(frac{-1}{4}right)+1 t &=frac{1}{4}-frac{2}{4}+1=frac{1}{4}+frac{1}{2}=frac{3}{4} therefore y &=frac{1}{frac{3}{4}}=frac{4}{3} end{aligned} )

# Q.6 The maximum value of the function y=7,2017 18-

Solution