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Algebraic Expressions Exercises

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1. Find the absolute extrema of the function π‘“αˆΊπ‘₯ሻ = 3π‘₯ 4 βˆ’ 4π‘₯ 3 on the interval αˆΎβˆ’1,2ሿ To find extreme points, we find and set 𝑓 β€² ሺπ‘₯ሻ to 0: 𝑓 β€² ሺπ‘₯ሻ = 12π‘₯ 3 βˆ’ 12π‘₯ 2 = 12π‘₯ 2 ሺπ‘₯ βˆ’ 1ሻ 2 𝑓 β€² ሺπ‘₯ሻ = 0 β†’ 12π‘₯ 2 ሺπ‘₯ βˆ’ 1ሻ = 0 β†’ {12π‘₯ = 0 β†’ π‘₯ = 0 π‘₯βˆ’1=0β†’π‘₯ =1 Both points lie on the interval αˆΎβˆ’1,2ሿ π‘“αˆΊβˆ’1ሻ = 3αˆΊβˆ’1ሻ4 βˆ’ 4αˆΊβˆ’1ሻ3 = 7 Finding π‘“αˆΊπ‘₯ሻ for endpoints: { π‘“αˆΊ2ሻ = 3ሺ2ሻ4 βˆ’ 4ሺ2ሻ3 = 16 Finding π‘“αˆΊπ‘₯ሻ for extreme points: { ∴{ π΄π‘π‘ π‘œπ‘™π‘’π‘‘π‘’ π‘€π‘Žπ‘₯: ሺ2,16ሻ π΄π‘π‘ π‘œπ‘™π‘’π‘‘π‘’ π‘šπ‘–π‘›: ሺ1, βˆ’1ሻ π‘“αˆΊ0ሻ = 3ሺ0ሻ4 βˆ’ 4ሺ0ሻ3 = 0 π‘“αˆΊ1ሻ = 3ሺ1ሻ4 βˆ’ 4ሺ1ሻ3 = βˆ’1 3 2. Verify directly that 𝐹ሺπ‘₯ሻ = π‘₯ 2 𝑒 π‘₯ , πΊαˆΊπ‘’αˆ» = 𝑒2 ln 𝑒 + 𝑒2 , π‘Žπ‘›π‘‘ π»αˆΊπ‘€αˆ» = ξ𝑀 4 + 2𝑀 are the 3 antiderivatives of π‘“αˆΊπ‘₯ሻ = π‘₯ሺ2 + 3π‘₯ 3 αˆ»π‘’ π‘₯ , π‘”αˆΊπ‘’αˆ» = π‘’αˆΊ3 + 2 ln π‘’αˆ», π‘Žπ‘›π‘‘ β„ŽαˆΊπ‘€αˆ» = 2𝑀 3 +1 ξ𝑀 4 +2𝑀 , respectively. To test, we’ll find the derivative of 𝐹ሺπ‘₯ሻ, πΊαˆΊπ‘’αˆ», π‘Žπ‘›π‘‘ π»αˆΊπ‘€αˆ». If their derivatives match π‘“αˆΊπ‘₯ሻ, π‘”αˆΊπ‘’αˆ», π‘Žπ‘›π‘‘ β„ŽαˆΊπ‘€αˆ» respectively, then we have verified directly that 𝐹ሺπ‘₯ሻ, πΊαˆΊπ‘’αˆ», π‘Žπ‘›π‘‘ π»αˆΊπ‘€αˆ» are the antiderivatives of π‘“αˆΊπ‘₯ሻ, π‘”αˆΊπ‘’ ...
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