图中显示点 $A\big(1\frac12, 5\frac12\big)$ 和 $B\big(7\frac12, 3\frac12\big)$ 位于方程为 $y = 9x - (2x + 1)^{\frac32}$ 的曲线上。

(a) 求曲线最高点的坐标。

(b) 验证直线 $AB$ 是曲线在 $A$ 处的法线。

(c) 求阴影区域的面积。

先令导数等于零，再检查左右两侧的符号。求得横坐标后回代原式。

Differentiation gives $y'=9-3\sqrt{2x+1}$. Setting it to zero gives $2x+1=9$, or $x=4$. The derivative is positive to the left and negative to the right within the domain. The height is $9(4)-9^{3/2}=9$, so the maximum point is $(4,9)$.

分别求曲线在 $A$ 的切线斜率和 $AB$ 的斜率。两者相乘应为 $-1$。

At $A$, $y'=9-3\sqrt4=3$. The slope of $AB$ is $(3.5-5.5)/(7.5-1.5)=-1/3$. Their product is $-1$, and $AB$ passes through $A$, so $AB$ is the normal at $A$.

先积分求曲线下方面积，再减去直线 $AB$ 下方的梯形面积。

The area below the curve between $x=3/2$ and $x=15/2$ is

The trapezium below $AB$ has width $6$ and parallel sides $5.5$ and $3.5$, so its area is $\frac12(5.5+3.5)(6)=27$. The shaded area is $44.6-27=17.6$.

也可以先求直线方程，再直接积分上下两条图像的高度差。

The line $AB$ has equation $y=6-x/3$. The curve is above it between $A$ and $B$, so the shaded area is