Question 1 Evaluating Exponential Functions. Use a calculator to evaluate the function at the indicated values. Round your answer to three decimals. ℎ ( 𝑥 ) = 𝑒 𝑥 ; ℎ ( 1 ) , ℎ ( 𝜋 ) , ℎ ( − 3 ) , ℎ ( 2 ) h(x)=e x ;h(1),h(π),h(−3),h( 2 ) Answer 1 Step 1 Given, ℎ ( 𝑥 ) = 𝑒 𝑥 h(x)=e x . We use a calculator to obtain the values of ℎ ( 1 ) , ℎ ( 𝜋 ) , ℎ ( − 3 ) , ℎ ( 2 ) h(1),h(π),h(−3),h( 2 ). Value of ℎ ( 𝑥 ) h(x) Exponential Function Output ℎ ( 1 ) h(1) 𝑒 1 e 1 2.718 ℎ ( 𝜋 ) h(π) 𝑒 𝜋 e π 23.141 ℎ ( − 3 ) h(−3) 𝑒 − 3 e −3 0.049 ℎ ( 2 ) h( 2 ) 𝑒 2 e 2 4.113 Thus, ℎ ( 1 ) = 2.718 , ℎ ( 𝜋 ) = 23.141 , ℎ ( − 3 ) = 0.049 , ℎ ( 2 ) = 4.113 h(1)=2.718,h(π)=23.141,h(−3)=0.049,h( 2 )=4.113 Question 2 Evaluating Exponential Functions. Use a calculator to evaluate the function at the indicated values. Round your answer to three decimals. 𝑔 ( 𝑧 ) = ( 1 3 ) 𝑧 + 1 ; 𝑔 ( 1 2 ) , 𝑔 ( 2 ) , 𝑔 ( − 3.5 ) , 𝑔 ( − 1.4 ) g(z)=( 3 1 ) z+1 ;g( 2 1 ),g( 2 ),g(−3.5),g(−1.4) Answer 1 Step 1 Given, 𝑔 ( 𝑧 ) = ( 1 3 ) 𝑧 + 1 g(z)=( 3 1 ) z+1 . We use a calculator to obtain the values of 𝑔 ( 1 2 ) , 𝑔 ( 2 ) , 𝑔 ( − 3.5 ) , 𝑔 ( − 1.4 ) g( 2 1 ),g( 2 ),g(−3.5),g(−1.4). Value of 𝑔 ( 𝑧 ) g(z) Exponential Function Output 𝑔 ( 1 2 ) g( 2 1 ) ( 1 3 ) 1.5 ( 3 1 ) 1.5 0.192 𝑔 ( 2 ) g( 2 ) ( 1 3 ) 2.414 ( 3 1 ) 2.414 0.070 𝑔 ( − 3.5 ) g(−3.5) ( 1 3 ) − 2.5 ( 3 1 ) −2.5 15.588 𝑔 ( − 1.4 ) g(−1.4) ( 1 3 ) − 0.4 ( 3 1 ) −0.4 1.551 Thus, 𝑔 ( 1 2 ) = 0.192 , 𝑔 ( 2 ) = 0.070 , 𝑔 ( − 3.5 ) = 15.588 , 𝑔 ( − 1.4 ) = 1.551 g( 2 1 )=0.192,g( 2 )=0.070,g(−3.5)=15.588,g(−1.4)=1.551 Question 3 The ideal body weight 𝑊 W for men (in kilograms) as a function of height ℎ h (in inches) is given by the following function: 𝑊 ( ℎ ) = 51 + 2.3 ( ℎ − 59 ) W(h)=51+2.3(h−59) a) What is the ideal weight of a 6-foot male? b) Express the height ℎ h as a function of weight 𝑊 W. Verify your answer by checking that 𝑊 ( ℎ ( 𝑊 ) ) = 𝑊 W(h(W))=W and ℎ ( 𝑊 ( ℎ ) ) = ℎ h(W(h))=h. c) What is the height of a male who is at his ideal weight of 80 kilograms? Answer 1 a) Ideal weight for a 6-foot male ℎ = 6 × 12 = 72 inches h=6×12=72 inches 𝑊 ( 72 ) = 51 + 2.3 ( 72 − 59 ) W(72)=51+2.3(72−59) = 51 + 2.3 × 13 = 51 + 29.9 = 80.9 kilograms =51+2.3×13=51+29.9=80.9 kilograms b) Expressing height ℎ h as a function of weight 𝑊 W Starting with: 𝑊 = 51 + 2.3 ( ℎ − 59 ) W=51+2.3(h−59) Solving for ℎ h: ℎ = 𝑊 − 51 2.3 + 59 h= 2.3 W−51 +59 Verification: 𝑊 ( ℎ ( 𝑊 ) ) = 𝑊 W(h(W))=W ℎ ( 𝑊 ( ℎ ) ) = ℎ h(W(h))=h c) Height of a male whose ideal weight is 80 kilograms ℎ ( 80 ) = 80 − 51 2.3 + 59 h(80)= 2.3 80−51 +59 = 29 2.3 + 59 ≈ 12.61 + 59 = 71.61 inches = 2.3 29 +59≈12.61+59=71.61 inches Question 4 Find any holes, vertical asymptotes, and horizontal asymptotes for the following rational function. 𝑓 ( 𝑥 ) = 2 ( 𝑥 − 3 ) ( 3 𝑥 + 5 ) 3 ( 3 𝑥 + 4 ) ( 𝑥 − 3 ) f(x)= 3(3x+4)(x−3) 2(x−3)(3x+5) Answer 1 Step 1 The function 𝑓 ( 𝑥 ) = 2 ( 𝑥 − 3 ) ( 3 𝑥 + 5 ) 3 ( 3 𝑥 + 4 ) ( 𝑥 − 3 ) f(x)= 3(3x+4)(x−3) 2(x−3)(3x+5) . Holes: The factor ( 𝑥 − 3 ) (x−3) cancels out, so there is a hole at 𝑥 = 3 x=3. Vertical Asymptote: The vertical asymptote occurs where the denominator is zero after cancelling common factors, so 𝑥 = − 4 3 x=− 3 4 . Horizontal Asymptote: The degrees of the numerator and denominator are the same (both degree 2), so the horizontal asymptote is the ratio of the leading coefficients: 𝑦 = 2 3 y= 3 2 Question 5 Write an equal decimal having the given number of decimal places. Briefly describe the Decimal Squares for both decimals in the equation. a. 0.7 = _ _ _ _ _ _ 0.7=______ (hundredths) Answer 1 Step 1 0.7 can be written as 7 10 = 70 100 10 7 = 100 70 . 7 10 10 7 means 7 out of 10 parts shaded. 70 100 100 70 means 70 out of 100 parts shaded. Answer: Option (A): 7 out of 10 parts shaded, or 70 out of 100 parts shaded.