1.3 and 3.9: Derivatives of exponential and logarithmic functions

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. and.9: Derivative of exponential and logarithmic function Problem Explain what each of the following mean: (a) f (x) Thi denote the invere function of f, f, evauluated at x. (b) f(x ) Thi mean f.

1 . and.9: Derivative of exponential and logarithmic function Problem Explain what each of the following mean: (a) f (x) Thi denote the invere function of f, f, evauluated at x. (b) f(x ) Thi mean f. x (c) (f(x)) Thi mean f(x) raied to the power, i.e. f(x). Problem 2 If f(x) repreent the number of package of bun needed for x package of hotdog, what doe f (x) repreent? f (x) repreent the number of packge of hotdog needed for x package of bun. Problem We re given the following graph of a function:

2 Ue thi graph to anwer the following quetion: (a) What i the domain of thi function? (, ) [ (, ) (b) What i the range of thi function? (, ) [ (0, ) (c) What i the value of f(0), f(), and f(2)? f(0) =, f() doe not exit, f(2) = 2 (d) Doe thi function have an invere? (Why or why not?) No, the function doe not have an invere. It i not one-to-one (that i, it doe not pa the horizontal line tet). (e) Find at leat two interval on which the function i one-to-one. The function become one-to-one when we retrict it domain to (, 0): The function alo become one-to-one when we retrict it domain to [0, ): 2

3 (f) Find f () on a retricted domain of f. In thi cae retrict the domain of f to [0, ). By definition we have f () = y () = f(y). Looking at the econd graph of the retricted function we ee that f(0) =, that i, f () = 0. (If we retric the domain of f to (, 0), thenf () = y () =f(y) implie that f(.7). Hence.7 f ().) (g) Uing the retricted domain for f of (, 0), ketch a graph of f. To graph the invere, we can think of the graph being reflected over the line y = x. Another way to obtain the graph i to remember than if (x, y) i a point on the graph of f, then(y, x) i a point on the graph of f. For example, (, ) i on the graph of f o (, ) i on the graph of f. (h) Uing the retricted domain for f of [0, ), ketch a graph of f. We can graph the invere of f on [0, ) in piece from left to right. Since (0, ) i a point on the graph of f, the point (, 0) in on the graph of f. Then, we ee on the graph of f, thereia linear piece going from (0, ) to (, 0). When we imagine reflecting thi part of the graph of f over the line y = x, it reflect onto itelf. The lat piece of the graph of f i a line from (, ) which appear to alo contain the point ( 2, 2). Reflecting thi part of f over the line y = x, we obtain a line tarting at (, ) and through the point ( 2, 2).

4 Problem 4 Let g be a one-to-one function and let g be it invere. True or Fale: If the point (2, /5) lie on the graph of g, then the point (2, 5) lie on the graph of g. Thi tatement i fale: wehaveg(2) = /5 () 2=g (/5). The notation g never, in thi coure, mean /g. Problem 5 Each of the following function are invertible on their given domain. For each one find a formula for it invere and give the domain and range of the invere. (a) The function f defined by f(x) =x 2 4x 5 for every x 2. To help find the a formula for f we will firt complete the quare : x 2 4x 5=x 2 4x , =(x 2) 2 9. Setting y = f(x) =x 2 4x 5=(x 2) 2 9, we can follow the procedure outlined for algebracially finding the formula for an invere function. y =(x 2) 2 9 =) y +9=(x 2) 2 =) p y +9= x 2 =) p y +9=x 2 (ince x 2) =) p y +9+2=x =) 2+ p x +9=y (interchange x and y along with minor rewriting) Therefore we have that f i defined by f (x) =2+ p x +9. The domain of f (x) i [ the range i [2, ). 9, ) and (b) The function g defined by g(u) = 4p u +2. Following the procedure to algebraically find the formula for the invere function we have z = 4p u +2 =) z =(u + 2) /4 =) z 4 = u +2 =) z 4 2=u =) u 4 2=z (interchange u and z) Therefore we have that g i defined by g (u) =u 4 i [ 2, ). 2. The domain of g i [0, ) and the range (c) The function h defined by h(t) =/(t + 2) 2 for every t> 2. 4

5 Following the procedure to algebraically find the formula for the invere function we have = (t + 2) 2 =) (t + 2) 2 = r =) t +2 = r =) t +2= r =) t = 2 (ince t> 2) =) = p t 2 (interchange and t) Therefore we have h i defined by h = p t 2. The domain of h i (0, ) and the range i ( 2, ). (d) The function p defined by p() =e +. Following the procedure to algebraically find the formula for the invere function we have y = e + =) ln(y) = + (ince ln i the invere of the natural exponential function) =) ln(y) = =) = ln(y) =) y = ln() (interchange y and ) Therefore we have p i defined by p () =(ln() i (, ). )/. The domain of p i (0, ) and the range Problem 6 Find all real number x which atify each of the following equation. (a) log x 25 = 2. Recall that, by definition of the invere to an exponential function, log b x = y () x = b y. Uing thi relationhip we have log x 25 = 2 () x 2 = 25. Therefore x 2 = 25 =) x = ±5, =) x =5 (bae of log i alway > 0). Therefore x =5i the only olution to log x 25 = 2. 5

6 (b) 7 x = 5 Similar to the previou problem, log b x = y () x = b y. Uing thi relationhip we have 7 x = 5 () x = log 7 5. Therefore x = log 7 5 i the only olution to 7 x = 5. (c) ln(x)+=0. Similar to the previou two problem we ll ue the relationhip ln(x) =y () x = e y. Before applying thi relationhip we perform a bit of algebra firt: ln(x)+=0 =) ln(x) =. Now we have ln(x) = () x = e. Therefore x = e (= /e) i the only olution to ln(x)+ = 0. Problem 7 True or Fale: () If f(x) =(x 2) x,thenf 0 (x) =x(x 2) x. Fale. Any time that you have a function of x raied to a function of x, in order to compute the derivative you need to ue logarithmic di erentiation (or omething equivalent). Correct derivative of f: (2) If f(x) =(x) x,thenf 0 (x) =(x) x ln(x). Fale. Same a part (). Correct derivative of f: f(x) =(x 2) x x ln(x 2) =) f(x) =e =) f 0 (x) =e x ln(x 2) ln(x 2) + x x 2 =) f 0 (x) =(x 2) x ln(x 2) + x x 2 f(x) =(x) x =) f(x) =e x ln(x) =) f 0 (x) =e x ln(x) ln(x)+x =) f 0 (x) =(x) x (ln(x) + ) x Problem 8 Find the derivative of the following function: 6

7 (a) f(x) =x ex +7x f 0 (x) = d x ex + d (7x) = d x ex +7. So the real problem i to find d x ex. Thu, f 0 (x) =x ex e x ln x + ex +7. x d x ex = d e ln xex = d e ex ln x = e ex ln x e x ln x + ex x = x ex e x ln x + ex x. (b) g(x) =(lnx + 9) ec(x4 ) g 0 (x) = d (ln x + 9) ec(x4 ) = d e ec(x4 )ln(lnx+9) = e ec(x4 )ln(lnx+9) 4x ec(x 4 ) tan(x 4 )ln(lnx + 9) + ec(x 4 ) x ln x +9 =(lnx + 9) ec(x4 ) 4x ec(x 4 ) tan(x 4 )ln(lnx + 9) + ec(x4 ). x(ln x + 9) (c) h(x) = (x2 7) 5 co 7 (x 2 5) Rewrite h(x) uing propertie of logarithm: (x 2 7) 5 ln h(x) =ln co 7 (x 2 5) =5 ln(x 2 7) 7 ln(co(x 2 5)) Derivative of h: d ln h(x) =) h0 (x) h(x) =5 x 2 7 2x 7 co(x 2 5) in(x2 5) 2x = 0x x x in(x2 5) co(x 2 5) = 0x x x tan(x2 5) 0x =) h 0 (x) =h(x) = h 0 (x) = (x2 7) 5 co 7 (x 2 5) x x tan(x2 5) 0x x x tan(x2 5) 7

MATEMATIK Datum: Tid: eftermiddag. A.Heintz Telefonvakt: Anders Martinsson Tel.:

MATEMATIK Datum: Tid: eftermiddag. A.Heintz Telefonvakt: Anders Martinsson Tel.: MATEMATIK Datum: 20-08-25 Tid: eftermiddag GU, Chalmer Hjälpmedel: inga A.Heintz Telefonvakt: Ander Martinon Tel.: 073-07926. Löningar till tenta i ODE och matematik modellering, MMG5, MVE6. Define what