Find the equation of the tangent to the curve y=e^x at the point where x=-1.
Equation for the tangent :
Delta y/Delta x = y'(-1)------>
[y-exp(-1)]/(x+1) = exp(-1) ------>
y = exp(-1) + (x+1)exp(-1) =
exp(-1)(2 + x)
I really don't understand this. Can you explain it step by step? The answer at the back of the book is y=-ex.
to find the equation of the tangent you will need the point of contact and the slope at that point of contact.
when x= -1, y = e^-1 = 1/e
dy/dx = e^x, so when x = -1 the slope = e^-1 = 1/e
using y = mx + b
1/e = (1/e)(-1) + b
2/e = b
the equation of the tangent is
y = (1/e)x + 2/e
To find the equation of the tangent to the curve y = e^x at the point where x = -1, we can use the method of finding the slope of the tangent and the point-slope form of a line.
Step 1: Find the slope of the tangent line.
The slope of the tangent line is equal to the derivative of the function at the given point. In this case, we need to find the derivative of y = e^x and evaluate it at x = -1.
The derivative of y = e^x with respect to x can be found using the power rule for differentiation, which states that d/dx (e^x) = e^x. Therefore, the derivative of y = e^x is dy/dx = e^x.
Evaluating the derivative at x = -1, we have dy/dx = e^(-1) = 1/e.
Step 2: Find the point of tangency.
To find the point of tangency, we substitute x = -1 into the original equation y = e^x.
So, when x = -1, we have y = e^(-1) = 1/e.
Therefore, the point of tangency is (-1, 1/e).
Step 3: Write the equation of the tangent line.
Using the point-slope form of a line, we have the equation:
y - y1 = m(x - x1),
where (x1, y1) represents the point of tangency and m represents the slope of the tangent.
Substituting the values into the equation, we get:
y - (1/e) = (1/e)(x - (-1)).
Simplifying this equation, we have:
y - (1/e) = (1/e)(x + 1).
Hence, the equation of the tangent to the curve y = e^x at the point where x = -1 is:
y - (1/e) = (1/e)(x + 1).