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