fx5ex AFind the slope of the graph of fx at the point where

f(x)=5-e^x

A.Find the slope of the graph of f(x) at the point where the graph crosses the x-axis.
slope = ??

B. Find the equation of the tangent line to the curve at this point.
y=????

C.ind the equation of the line perpendicular to the tangent line at this point. (This is the normal line.)
y=???

Solution

f(x) = 5 - e^x

f\'(x) = -e^x

(A) For crossing the x-axis, f(x) = 0

5 - e^x = 0

e^x = 5

x = ln5

Slope at x = ln5, f \' (ln5) = -e^(ln5) = -5

(B) At x = ln5, y = 0

Slope = -5

Equation of tangent :

y - 0 = -5 (x - ln5)

y = -5x + 5ln5

5x + y = 5 ln5 (Tangent)

(C) Slope of the normal, m = -1/f \' (x)

At x = ln5, f\'(x) = -5

slope of normal = 1/5

Equation of normal :

y - 0 = (1/5)(x - ln5)

5y = x - ln5

x - 5y = ln5 (Normal)

f(x)=5-e^x A.Find the slope of the graph of f(x) at the point where the graph crosses the x-axis. slope = ?? B. Find the equation of the tangent line to the cur

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