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10. Consider the function: f(r) = ¹-. Find all maxima, minima and points of inflection for this function. Plot the function.

10. Consider the function: f(r) = ¹-. Find all maxima, minima and points of inflection for this function. Plot the function.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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1) 2 ~18165 10. Consider the function: f(x)=r¹-6. Find all maxima, minima and points of inflection for this function. Plot the function. 6 f(x) = x²-x² Of = R 0096 | Rf = (-00, max 4 f(x) | 0 3 3 f'(x) = 4x² - 6x² = x² (1-62²) The function has inflection points at. x= + √5 +₁ ( √753) = ( 3 ) ² (1-4/2)) > 0 1096 and stays positive around. √575 x=0 We need to check x=0₁ min for max or and x=0 ƒ" (√¾/5) = − ( 2 ) ^ (1-669) - <० fl=0 => fill= 122² - 30x² =6x2² (2-5x²) =0 if x = 0, 6x=+√√¾/5 .8165 and starp negatiul aroud -√2/15 2² 4-6x² = 0 = x = ± √√√²/₂ x= ± √575 for in flecter points. 2 x x² (1-x²) Soi f"(x) < 0 at x = 18165 0 f"(x) <0 at x==18165 11 At x = 0 fhl=0 at x=0 x= ± √ √73 So we continul. f (x) = 24x - 120x² f(x) = 24-240x O x = +1 +.8165 =+6325 both are maxima taking descontives But f" (0) = 2470 we have an even derivative as the first non-zero derivatue and it is positive. x=0 is a minimum
Transcribed Image Text:1) 2 ~18165 10. Consider the function: f(x)=r¹-6. Find all maxima, minima and points of inflection for this function. Plot the function. 6 f(x) = x²-x² Of = R 0096 | Rf = (-00, max 4 f(x) | 0 3 3 f'(x) = 4x² - 6x² = x² (1-62²) The function has inflection points at. x= + √5 +₁ ( √753) = ( 3 ) ² (1-4/2)) > 0 1096 and stays positive around. √575 x=0 We need to check x=0₁ min for max or and x=0 ƒ" (√¾/5) = − ( 2 ) ^ (1-669) - <० fl=0 => fill= 122² - 30x² =6x2² (2-5x²) =0 if x = 0, 6x=+√√¾/5 .8165 and starp negatiul aroud -√2/15 2² 4-6x² = 0 = x = ± √√√²/₂ x= ± √575 for in flecter points. 2 x x² (1-x²) Soi f"(x) < 0 at x = 18165 0 f"(x) <0 at x==18165 11 At x = 0 fhl=0 at x=0 x= ± √ √73 So we continul. f (x) = 24x - 120x² f(x) = 24-240x O x = +1 +.8165 =+6325 both are maxima taking descontives But f" (0) = 2470 we have an even derivative as the first non-zero derivatue and it is positive. x=0 is a minimum
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