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In Exercises 4–10, find the global maximum and minimum for the function on the closed interval.
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Calculus: Single And Multivariable
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- In Exercises 33–38, express the function, f, in simplified form. Assume that x can be any real number. 33. f(x) = V36(x + 2)² 34. f(x) = V81(x – 2)2 35. f(x) = V32(x + 2)³ 36. f(x) = V48(x – 2)³ 37. f(x) = V3x² – 6x + 3 38. f(x) = V5x2 – 10x + 5 %3Darrow_forwardExercises 121–140: (Refer to Examples 12–14.) Complete the following for the given f(x). (a) Find f(x + h). (b) Find the difference quotient of f and simplify. 121. f(x) = 3 122. f(x) = -5 123. f(x) = 2x + 1 124. f(x) = -3x + 4 %3D 125. f(x) = 4x + 3 126. f(x) = 5x – 6 127. f(x) = -6x² - x + 4 128. f(x) = x² + 4x 129. f(x) = 1 – x² 130. f(x) = 3x² 131. f(x) = 132. /(x) 3D글 = = 132. f(: 133. f(x) = 3x² + 1 134. f(x) = x² –- 2 135. f(x) = -x² + 2r 136. f(x) = -4xr² + 1 137. f(x) = 2x - x +1 138. f(x) = x² + 3x - 2 139. f(x) = x' 140. f(x) = 1 – xarrow_forwardIn Exercises 57–62, find the zeros of ƒ and sketch its graph by plotting points. Use symmetry and increase/decrease information where appropriate. 57. f(x) — х? — 4 58. f(x) = 2x2 – 4 %3D %3D 59. f(x) — х3 — 4х 60. f(x) — х3 61. f(x) =2 – x3 62. f(x) = (x – A)¾i+ate Windarrow_forward
- In Exercises 12–20, find all zeros of each polynomial function. Then graph the function. 12. f(x) = (x – 2)°(x + 1)³ 13. f(x) = -(x – 2)(x + 1)? 14. f(x) = x - xr? – 4x + 4 15. f(x) = x* - 5x² + 4 16. f(x) = -(x + 1)° 17. f(x) = -6x³ + 7x? - 1 18. f(x) = 2r³ – 2x 19. f(x) = x - 2x² + 26x 20. f(x) = -x + 5x² – 5x – 3 %3D %3D %3! %3D %3!arrow_forwardThe critical point of the function f(x) = x³ – x? + 2x –23 is (을,2) (금, -2) | (응, -2) (금,2)arrow_forwardThe absolute maximum and absolute minimum value of the function f(x) = 3x - in [-1,4] is: 4.5, 4 -4.5, -4 -4.5, -3.5 4.5, -3.5arrow_forward
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- Pls help ASAParrow_forward1. Let g: Z→→ Z be the function g(x)=3-2x and let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Determine g(A). Explain your reasoning. (Video 2.4A)arrow_forwardEXERCISE 6 (PSAT – 20151028 Section 4, Q9) (Calculator permitted) f(x) = x + 4(x – 3) Which of the following is an equivalent form of the expression above that displays the zeros of the function as constants or coefficients in the expression? A) f(x) = x² + 4x – 12 B) f(x) = (x – 6)(x + 2) C) f(x) = (x + 6)(x – 2) D) f(x) = (x +2)² – 16arrow_forward
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