Math Problem Statement

ΰΆ± 0 π‘₯ 𝑓𝑑𝑑𝑑=βˆ’1 2+π‘₯2+π‘₯𝑠𝑒𝑛2π‘₯ +1 2 cos(2π‘₯) 𝑓(1 4πœ‹) 𝑓′(1 4πœ‹) 𝑓(1 4πœ‹) βˆ’1 2+ 1 4πœ‹ 2 + 1 4πœ‹ 𝑠𝑒𝑛 21 4πœ‹ +1 2 cos(21 4πœ‹) βˆ’1 2+1 16πœ‹2+1 4πœ‹π‘ π‘’π‘› πœ‹ 2 +1 2 cos(πœ‹ 2) βˆ’1 2+1 16πœ‹2+1 4πœ‹1+1 2 (0) βˆ’1 2+1 16πœ‹2+1 4πœ‹ 𝑓′(1 4πœ‹) 2π‘₯+𝑠𝑒𝑛2π‘₯ +2π‘₯π‘π‘œπ‘ 2π‘₯ βˆ’1 22𝑠𝑒𝑛(2π‘₯) 2π‘₯+𝑠𝑒𝑛2π‘₯ +2π‘₯π‘π‘œπ‘ 2π‘₯ βˆ’π‘ π‘’π‘›(2π‘₯) 1 1 1 1 1 1 2 1 1 4πœ‹ +𝑠𝑒𝑛 2πœ‹+𝑠𝑒𝑛 2 4πœ‹ +2 πœ‹ 2 +πœ‹ 2 cos Problema 4 Hallar todos los valores de x π‘₯ ΰΆ± 0 𝑑3 βˆ’π‘‘ 𝑑𝑑 =1 3ΰΆ± 1 4πœ‹cos πœ‹ πœ‹ 2 βˆ’π‘ π‘’π‘›( 𝑑 βˆ’π‘‘3 𝑑𝑑 2 οΏ½ οΏ½ 1 π‘₯4 βˆ’2π‘₯2 4 = 3 2π‘₯2 βˆ’π‘₯4 4 βˆ’ 2 ) 21 4πœ‹ βˆ’π‘ π‘’π‘›(2 4 2 2 βˆ’ 4 0 1 4πœ‹) πœ‹ 2 +1+0βˆ’1 1 π‘₯4 4 βˆ’ π‘₯2 2 = 3 3 πœ‹ 2 π‘₯2 2 βˆ’ π‘₯4 4 βˆ’ es correcto?

Solution

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Trigonometry
Derivatives

Formulas

f(x) = -1 / (2 + x^2 + x*sin^2(x) + 1/2*cos(2x))
Derivative of f(x)
Integral of f(x) from 0 to x

Theorems

Quotient Rule for Derivatives
Trigonometric Identities

Suitable Grade Level

University Level