Find an equation of the tangent line to the graph of y = e^(−x^2) at the point (5, 1/(e^25)).
My work:
y' = -2xe^(-x^2)
y' when x = 5 = -10e^25
y - (1/e^25) = -10e^25 * (x - 5)
y = (-10e^25) (x - 5) + (1/e^25)
This is incorrect. How?
y' when x = 5 = -10e^-25
To find the equation of the tangent line to the graph of y = e^(-x^2) at the point (5, 1/(e^25)), you've correctly found the derivative of the function y = e^(-x^2). However, there seems to be a mistake when substituting the value of x into the derivative.
Let's calculate the value of y' when x = 5 again:
y' = -2xe^(-x^2)
y' when x = 5 = -2(5)e^(-5^2)
= -10e^(-25)
= -10/e^25
Notice that we need to evaluate e^(-25), not e^25. The negative exponent means we're looking for the reciprocal of e^25, not raising e to the power of 25.
So, the correct equation for the tangent line is:
y - (1/e^25) = (-10/e^25) * (x - 5)
To simplify further, we can multiply both sides of the equation by e^25 to eliminate the fractions:
e^25 * y - 1 = -10(x - 5)
Finally, we can rewrite the equation in slope-intercept form:
e^25 * y = -10x + 50 - 1
y = (-10/e^25)x + (49/e^25)
Therefore, the correct equation of the tangent line to the graph of y = e^(-x^2) at the point (5, 1/(e^25)) is y = (-10/e^25)x + (49/e^25).