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Milana03091994
@Milana03091994
July 2022
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sangers1959
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1) f(x)=log₀,₅(2+x)
f`(1)=1/(ln0,5*(2+x))=1/(3*ln0,5)<0, так как ln0,5<0≈-0,7 (0,5<e).
2) f(x)=log₃(5+x)
f`(4)=1/(ln3*(5+x))=1/(9*ln3)>0 (3>e).
3)f(x)=0,2⁽ˣ⁻³⁾
f`(4)=ln0,2*(0,2)⁽⁴⁻³⁾=ln0,2*(1/5)¹=ln0,2/5<0 (0,2<e).
4) f(x)=2,5⁽ˣ⁻¹⁾
f`(2)=ln2,5*2,5¹=2,5*ln2,5<0 (2,5<e)
e≈2,72.
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Answers & Comments
Verified answer
1) f(x)=log₀,₅(2+x)f`(1)=1/(ln0,5*(2+x))=1/(3*ln0,5)<0, так как ln0,5<0≈-0,7 (0,5<e).
2) f(x)=log₃(5+x)
f`(4)=1/(ln3*(5+x))=1/(9*ln3)>0 (3>e).
3)f(x)=0,2⁽ˣ⁻³⁾
f`(4)=ln0,2*(0,2)⁽⁴⁻³⁾=ln0,2*(1/5)¹=ln0,2/5<0 (0,2<e).
4) f(x)=2,5⁽ˣ⁻¹⁾
f`(2)=ln2,5*2,5¹=2,5*ln2,5<0 (2,5<e) e≈2,72.