- #1
hvroegindewey
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Does anyone know how to find the equation of a tangent line to y=e^(-2x) at the point (1,e^-2) I honestly have no idea how to even start this problem when I tried it I came up with y=1(x+.27)+ln2
The purpose of solving for the tangent line at (1,e^-2) is to find the slope of the line at that specific point on the curve. This can provide valuable information about the behavior of the curve and can be used to make predictions and calculations.
The steps involved in solving for the tangent line at (1,e^-2) include finding the derivative of the curve, plugging in the x-coordinate of the given point into the derivative to find the slope, and using the point-slope formula to write the equation of the tangent line.
Solving for the tangent line at (1,e^-2) is important because it allows us to approximate the behavior of the curve at that specific point. This information can be used in various applications, such as in physics and engineering, to make predictions and solve problems.
The point (1,e^-2) is significant because it represents a specific point on the curve at which we want to find the tangent line. This point can also provide information about the overall behavior of the curve, such as its concavity and rate of change.
Yes, the process of solving for the tangent line at (1,e^-2) can be applied to any curve, as long as the curve is differentiable at that point. This means that the curve must have a well-defined slope at the given point.