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The minimum value of `y=5x^(2)-2x+1` isA. `(1)/(5)`B. `(2)/(5)`C. `(4)/(5)`D. `(3)/(5)` |
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Answer» Correct Answer - C The minimum/minimum value `(dy)/(dx)=0rArr5(2x)-2(1)+0=0=0rArrx=(1)/(5)` Now at `x=(1)/(5)m(d_(y^2))/(dx^(2))=10 x=(1)/(5)m(d_(y^2))/(dx^(2))=10` which is positive so minima at `x=(1)/(5)`. Therefore `y_("min")=5((1)/(5))^(2)-2((1)/(5))+1=(4)/(5)` |
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