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### Transcript of Calc 5.4a

• 1. DEVELOP PROPERTIES OF THE NATURAL EXPONENT FUNCTION DIFFERENTIATE NATURAL EXPONENTIAL FUNCTIONS 5.4aExponential Functions:Differentiation

2. 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= xand e ln x= x 3. Ex 1 p. 350Solving 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, rewritein logarithmic form: 4. Ex 2 p.351Solving 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! 5. 6. The coolest thing aboute xis that it is its own derivative! Proof:We know thatln e x=x.So if we take the derivative of both sides,Multiply both sides by e x 7. A geometrical interpretation is that the slope of the graph of f(x) =e xat any point (x,e x ) is equal to the y-value of the point. Ex 3 p. 352Differentiating Exponential Functions u = 3x-2 u= -2x -1 8. Ex 4, p352 Locating Relative Extrema Find the relative extrema ofUsing 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. 9. Ex 5 p. 353The Standard Normal Probability Density Function Show that thestandard 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, theyarepoints of inflection when x = 1 10. 5.4a p. 356/ 1-61 every other odd