Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
At x = 0, y = 2.
At y = 16, x = 1.
Gradient of the curve at x = 1 or at the tangent is;
dy/dx = 3In2*2^(3x+1)
For x = 1
dy/dx = 48In2
Therefore coordinate Q is (x¹, 0).
Calculating x¹.
(0-16)/(x¹-1) = 48In2
x¹= (3In2-1)/3In2
It implies;
Area Red exactly as a single fraction in its simplest form is;
Area trapezoid with parallel sides 2 units and 16 units, and height 1 unit - Area triangle with height 16 units and base (1-((3In2-1)/3In2)) units.
= (½*1(2+16)) - (½(16)(1-((3In2-1)/3In2))
= 9 - (8(1/3In2))
= 9 - (8/(3In2))
= (27In2-8)/3In2 space units.
= 5.15281322429 space units.
