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C4 Maths Question....?
Ok, got another attempt at a question and need to know if i've failed, got it right, got the rigt working etc.
So here it is.
Show that the equation of the tangent to y = e^-3x at the point (0,1) is y = 1 - 3x
part B is
Find the area bounded by y = e^-3x, the axes and x= -1
Man C4 is hard :/
1 Answer
- Ron WLv 71 decade agoFavorite Answer
The slope of the tangent is the value of the derivative at the point of tangency.
y(x) = e^(-3x)
y'(x) = -3e^(-3x)
y'(0) = -3
Point-slope form of the equation of a line: y - y0 = m(x - x0) where m is the slope and (x0, y0) is on the line. Thus, the equation of the tangent is
y - 1 = -3(x - 0)
which simplifies to y = -3x + 1 (or 1 - 3x)
part B
Did you sketch the region?
It is bounded on the bottom by the x-axis (whose equation is y = 0), on the left by the line x = -1, on the right by the y-axis (whose equation is x = 0), and on the top by the curve y = e^(-3x)
......0
A = ∫ [e^(-3x) - 0] dx = (-1/3)e^(-3x) (evaluated at -1 and 0)
.....-1
= (-1/3)( e^[(-3)(0)] - e^[(-3)(-1)] ) = (-1/3)(1 - e^3) = (e^3 - 1)/3
