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The least value of the function $f(x)=x^2+x+2$ in the interval [0, 1] is: |
4 $\frac{7}{4}$ $\frac{3}{4}$ 2 |
2 |
The correct answer is Option (2) → $\frac{7}{4}$ $f(x)=x^2+x+2$ $f'(x)=2x+1$ $2x+1=0 \Rightarrow x=-\frac{1}{2} \notin [0,1]$ $f(0)=2,\quad f(1)=4$ $\text{Least value} = 2$ $\text{Minimum value on } [0,1] = 2$ |
