設(shè)a為實(shí)數(shù),函數(shù)f(x)=e的x方-2x+2a x屬于R 求f(x)的單調(diào)區(qū)間與極值求證當(dāng)a大于ln2-1且x大于0時(shí),e的x方...
設(shè)a為實(shí)數(shù),函數(shù)f(x)=e的x方-2x+2a x屬于R 求f(x)的單調(diào)區(qū)間與極值求證當(dāng)a大于ln2-1且x大于0時(shí),e的x方大于x的平方2-2ax+1
f(x)=e的x方-2x+2a
f'(x)=e的x方-2=0
x=ln2
e^x是增函數(shù),所以
當(dāng)x>ln2時(shí),f'(x)>0
當(dāng)x
所以
增區(qū)間為:(ln2,+∞)
減區(qū)間為:(-∞,ln2)
從而
當(dāng)x=ln2時(shí)又極小值f(ln2)=2-2ln2+2a