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Question Number 185677 by Mingma last updated on 25/Jan/23

Answered by Frix last updated on 25/Jan/23

(df/dx)=e^(∣x∣) −e^(∣x−2∣) =0 ⇒ x=1 f(1)=∫_0 ^2 e^(∣1−t∣) dt=2∫_0 ^1 e^(1−t) dt=2[−e^(1−t) ]_0 ^1 = =2(e−1)

dfdx=e∣x∣−e∣x−2∣=0⇒x=1f(1)=∫20e∣1−t∣dt=2∫10e1−tdt=2[−e1−t]01==2(e−1)

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