Hello, gracey!
Given: .$\displaystyle y\:=\: (2x3)^3$
(a) Find the xcoordinates of the two points on the curve which have gradient 6.
We have: .$\displaystyle y' \:=\:6(2x3)^2 \:=\:6 \quad\Rightarrow\quad(2x3)^2 \:=\:1 \quad\Rightarrow\quad 2x3 \:=\:\pm1$
. . $\displaystyle \begin{array}{cccccc}2x3 \:=\:1 & \Rightarrow & 2x\:=\:4 & \Rightarrow & \boxed{x \:=\:2} \\ \\[3mm] 2x3\:=\:\text{}1 & \Rightarrow & 2x \:=\:2 & \Rightarrow & \boxed{x \:=\:1} \end{array}$
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(b) The region under the xaxis and to the right of the yaxis
is bounded by the curve. .What is the area of the region?
$\displaystyle \text{Area} \;=\;\int^{\frac{3}{2}}_0 \text{}\,(2x3)^3\,dx \;=\;\text{}\,\frac{1}{8}(2x3)^4\,\bigg]^{\frac{3}{2}}_0 \quad\hdots\quad\text{etc.} $