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Let f(x) = (x² − 3)eª. (x² − 3)e. The function has two critical points at x = -3, and x = 1. (a) Use the second derivative test to classify each critical point as local max or local min. (b) Find the global maximum value M and the global minimum value m of the function f(x) on the interval ï € [−4, −2]. (c) Consider the definite integral -2 -2 (x² — 3)e* dx. Use your result from part (b) f(x) dx = fe -4 -4 to find an underestimate and overestimate for the integral. (d) Find the right sum using n=2, i.e., RIGHT(2), for the integral -2 -2 [₁² f(x) dx = [ -4 -4 below. Round your answer to two decimal places. 2 (x² − 3)e dx. Enter this value in the blank

Let f(x) = (x² − 3)eª. (x² − 3)e. The function has two critical points at x = -3, and x = 1. (a) Use the second derivative test to classify each critical point as local max or local min. (b) Find the global maximum value M and the global minimum value m of the function f(x) on the interval ï € [−4, −2]. (c) Consider the definite integral -2 -2 (x² — 3)e* dx. Use your result from part (b) f(x) dx = fe -4 -4 to find an underestimate and overestimate for the integral. (d) Find the right sum using n=2, i.e., RIGHT(2), for the integral -2 -2 [₁² f(x) dx = [ -4 -4 below. Round your answer to two decimal places. 2 (x² − 3)e dx. Enter this value in the blank

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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Let ƒ(x) = (x² − 3)eª. The function has two critical points at X = -3, and x = 1. (a) Use the second derivative test to classify each critical point as local max or local min. (b) Find the global maximum value M and the global minimum value m of the function f(x) on the interval ï € [−4, −2]. (c) Consider the definite integral -2 f(x) dx = f(x) dx -2 F -4 to find an underestimate and overestimate for the integral. = (x² − 3)eº dx. Use your result from part (b) -4 (d) Find the right sum using n=2, i.e., RIGHT(2), for the integral -2 -2 [₁² (2₂²-3) e² -4 -4 below. Round your answer to two decimal places. dx. Enter this value in the blank
Transcribed Image Text:Let ƒ(x) = (x² − 3)eª. The function has two critical points at X = -3, and x = 1. (a) Use the second derivative test to classify each critical point as local max or local min. (b) Find the global maximum value M and the global minimum value m of the function f(x) on the interval ï € [−4, −2]. (c) Consider the definite integral -2 f(x) dx = f(x) dx -2 F -4 to find an underestimate and overestimate for the integral. = (x² − 3)eº dx. Use your result from part (b) -4 (d) Find the right sum using n=2, i.e., RIGHT(2), for the integral -2 -2 [₁² (2₂²-3) e² -4 -4 below. Round your answer to two decimal places. dx. Enter this value in the blank
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