Lecture 3 : The Natural Exponential Function: f(x) = exp(x) = e x. y = exp(x) if and only if x = ln(y)

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1 Lecture 3 : The Natural Exponential Function: f(x) = exp(x) = Last day, we saw that the function f(x) = ln x is one-to-one, with domain (, ) and range (, ). We can conclude that f(x) has an inverse function f (x) = exp(x) which we call the natural exponential function. The definition of inverse functions gives us the following: The cancellation laws give us: y = f (x) if and only if x = f(y) y = exp(x) if and only if x = ln(y) f (f(x)) = x and f(f (x)) = x exp(ln x) = x and ln(exp(x)) = x. We can draw the graph of y = exp(x) by reflecting the graph of y = ln(x) in the line y = x. 5 y = exphxl = H, e L 5 H, el y = lnhxl H-7, e -7 L H, L He, L He, L -5 H, L 5-5 He -7, -7L - We have that the graph y = exp(x) is one-to-one and continuous with domain (, ) and range (, ). Note that exp(x) > for all values of x. We see that In fact for any rational number r, we have exp() = since ln = exp() = e since ln e =, exp() = e since ln(e ) =, exp( 7) = e 7 since ln(e 7 ) = 7. exp(r) = e r since ln(e r ) = r ln e = r,

2 by the laws of Logarithms. When x is rational or irrational, we define to be exp(x). = exp(x) Note: This agrees with definitions of given elsewhere, since the definition is the same when x is a rational number and the exponential function is continuous. Restating the above properties given above in light of this new interpretation of the exponential function, we get: = y if and only if We can use these formulas to solve equations. Example Solve for x if ln(x + ) = 5 ln y = x e ln x = x and ln = x Solving Equations Example Solve for x if = From the graph we see that x ex =, Limits =. x Example Find the it x.

3 Rules of Exponents The following rules of exponents follow from the rules of logarithms: +y = e y, y = ex e y, (ex ) y = y. Proof We have ln(+y ) = x + y = ln( ) + ln(e y ) = ln( e y ). Therefore +y = e y. The other rules can be proven similarly. Example Simplify ex e x+ ( ). d dx ex = Derivatives d dx eg(x) = g (x)e g(x) Proof We use logarithmic differentiation. If y =, we have ln y = x and differentiating, we get dy dy = or = y = y dx dx ex. The derivative on the right follows from the chain rule. Example Find d dx esin x and d dx sin ( ) dx = + C Integrals g (x)e g(x) dx = e g(x) + C Example Find x + dx. 3

4 Old Exam Questions Old Exam Question The function f(x) = x 3 + 3x + e x is one-to-one. Compute f( ) (). Old Exam Question Compute the it x e x e x e x. Old Exam Question Compute the Integral. ln + dx

5 Extra Examples (please attempt these before looking at the solutions) Example Find the domain of the function g(x) = 5. Example Solve for x if ln(ln(x )) = Example Let f(x) = e x+3, Show that f is a one-to-one function and find the formula for f (x). Example Evaluate the integral 3e ( 3e x ln x 3 ) 3 dx. Example Find the it x and x. Example Find π (cos x)esin x dx. Example Find the first and second derivatives of h(x) = ex. Sketch the graph of h(x) with horizontal, and vertical asymptotes, showing where the function is increasing and decreasing and showing intervals of concavity and convexity. 5

6 Extra Examples: Solutions Example Find the domain of the function g(x) = 5. The domain of g is {x 5 }. 5 if and only if 5 if and only if ln 5 ln( ) = x or x ln 5 since ln(x) is an increasing function. Example Solve for x if ln(ln(x )) = We apply the exponential function to both sides to get e ln(ln(x )) = e or ln(x ) = e. Applying the exponential function to both sides again, we get e ln(x) = e e or x = e e. Taking the square root of both sides, we get x = e e. Example Let f(x) = e x+3, Show that f is a one-to-one function and find the formula for f (x). We have the domain of f is all real numbers. To find a formula for f, we use the method given in lecture. y = e x+3 is the same as x = f (y). we solve for x in the equation on the left, first we apply the logarithm function to both sides ln(y) = ln(e x+3 ) = x + 3 x = ln(y) 3 x = ln(y) 3 Now we switch th and y to get y = ln(x) 3 = f (x). = f (y). Example Evaluate the integral 3e ( 3e x ln x 3 ) 3 dx. We try the substitution u = ln x 3. du = 3 x 3 dx = x dx, u(3e ) =, u(3e ) =. 3e ( 3e x ln x 3 ) 3 dx = u du = u3 6 = u

7 = Example Find the it x x x ( )(6) ( )() = 8 3 = 3 3 and x = =. x x ( ) = =. x x ( ) = = 9. Example Find π (cos x)esin x dx. We use substitution. Let u = sin x, then du = cos x dx, u() = and u(π/) =. π (cos x)e sin x dx = e u du = e u = e e = e. Example Find the first and second derivatives of h(x) = ex. Sketch the graph of h(x) with horizontal, and vertical asymptotes, showing where the function is increasing and decreasing and showing intervals of concavity and convexity. y-int: h() = 9 x-int: h(x) = if and only if =, this is impossible, so there is no x intercept. e H.A. : In class, we saw x x = and above, we saw x ex =. So the H.A. s are y = and y =. V.A. : The graph has a vertical asymptote at x if =, that is if = or x = ln( ). Inc/Dec (h (x)) To determine where the graph is increasing or decreasing, we calculate the derivative using the quotient rule h (x) = (ex ) ( ) ( ) = ex ( ) ( ) = ( ). Since h (x) is always negative, the graph of y = h(x) is always decreasing. Concave/Convex To determine intervals of concavity and convexity, we calculate the second derivative. h (x) = d dx h (x) = d dx ( ) = d dx ( ). I m going to use logarithmic differentiation here y = differentiating both sides, we get Multiplying across by y = ( ) ln(y) = ln( ) ln( ) = x ln( ) dy y dx = ex = ex. ( ), we get dy dx = ( ) ( ) 7.

8 = ex ( ) ( ) = ex ( ) = ex ( + ) ( ) 3 ( ) 3 ( ) 3 h (x) = dy dx = ex ( + ) ( ) 3 We see that the numerator is always positive here. From our calculations above, we have < if x < ln(/) and > if x > ln(/). Therefore h (x) < if x < ln(/) and h (x) > if x > ln(/) and The graph of y = h(x) is concave down if x < ln(/) and concave up if x > ln(/). Putting all of this together, you should get a graph that looks like: Check it and other functions out in Mthematica 8

9 Answers to Old Exam Questions Old Exam Question The function f(x) = x 3 + 3x + e x is one-to-one. Compute f( ) (). We use the formula (f ) () = f (f ()) b = f () same as f(b) = b 3 + 3b + e b =. Solving for b is very difficult, but we can work by trail and error. If we try b =, we see that it works, since e =. Therefore f () =. We also need to calculate f (x), we get f (x) = 3x e x. (f ) () = Old Exam Question Compute the it f (f ()) = f () = 3 + = 5. x e x e x e x. We divide both numerator and denominator by the highest power of in the denominator which is e x in this case. e x ( e x )/e x e x e 3x = = = x e x e x x (e x e x )/ex x e x =. Old Exam Question Compute the Integral. ln We make the substitution u = +. We have + dx du = dx, u() =, u(ln ) = + e ln = 3. We get ln 3 + e dx = x u du = ln u = ln 3 ln = ln(3/). 3 9