DEVELOP PROPERTIES OF THE NATURAL EXPONENT FUNCTION DIFFERENTIATE NATURAL EXPONENTIAL FUNCTIONS 5.4a Exponential Functions: Differentiation
f (x) = ln x is a function that is increasing on its entire domain, and so it has an inverse, f -1 (x) = e x f (x) f -1 (x) Since they are inverses, ln e x = x and e ln x = x
Ex 1 p. 350 Solving Exponential Equations Take the natural log of both sides (since base e) Another way to think of this is to put the exponential form into logarithmic form. From the original problem, to free up a variable in the exponential position, rewrite in logarithmic form:
Ex 2 p.351 Solving a logarithmic equation Exponentiate each side Another way to think of this is to put the log form back into exponential form – the base raised to the exponent(opp. side) is equal to the inside of the log. And so on!
The coolest thing about e x is that it is its own derivative! Proof: We know that ln e x =x. So if we take the derivative of both sides, Multiply both sides by e x
A geometrical interpretation is that the slope of the graph of f(x) = e x at any point (x, e x ) is equal to the y-value of the point. Ex 3 p. 352 Differentiating Exponential Functions u = 3x-2 u= -2x -1
Ex 4, p352 Locating Relative Extrema Find the relative extrema of Using the product rule, In the interval (-∞, -1) the derivative is negative so the function is decreasing. In the interval (-1, ∞), the derivative is positive, so the function is increasing. Since it switches from decreasing to increasing at x = -1, the point (-1, -2/e) is a relative minimum.
Ex 5 p. 353 The Standard Normal Probability Density Function Show that the standard normal probability density function has points of inflection at x = 1 Solution: to locate possible points of inflection, find the x-values for which the second derivative is 0 Setting f “(x) = 0 makes possible points of inflection happen at x = -1, 1 Testing intervals using Ch 3 concepts, they are points of inflection when x = 1