The number of real solution(s) of the equatio | vedclass

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(C) Let $f(x) = x^2+3x+2 = (x+1)(x+2)$.Let $g(x) = \min \{|x-3|, |x+2|\}$.We need to find the number of intersection points of $y = f(x)$ and $y = g(x

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$3$

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(C) Let $f(x) = x^2+3x+2 = (x+1)(x+2)$.Let $g(x) = \min \{|x-3|, |x+2|\}$.We need to find the number of intersection points of $y = f(x)$ and $y = g(x)$.For $x  0$ and $g(x) = |x-3| = 3-x$. Solving $x^2+3x+2 = 3-x$ gives $x^2+4x-1 = 0$,so $x = -2 \pm \sqrt{5}$. Since $x For $-2 \le x For $x \ge 0.5$,$g(x) = |x-3|$. The parabola $f(x)$ is increasing and $g(x)$ is decreasing or increasing,but $f(x)$ grows much faster,so there are no further solutions.The solutions are $x = -2-\sqrt{5}$,$x = -2$,and $x = 0$. Thus,there are $3$ real solutions.

The number of real solution$(s)$ of the equation $x^2+3x+2=\min \{|x-3|, |x+2|\}$ is:

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