Question: 1.>(40%) .True.and.False.1 1. = |sin(@)de = 0.9 - * 2 . f(x)=sin(x)-cos(x).has.1.as.a.local.maximum.because.both.sin.and. cos.have.1.as.a.local.maximum. . 1 3.- .If.a.function.is.differentiable, then.it.must be.continuous. 1 4.> The.total-areas.of.step.function,.f(k)=k.feet, .k.ranging from.1.to. 1000.is.500,500.ft2

1.>(40%) .True.and.False.1 1. = |sin(@)de = 0.9 - * 2 .

f(x)=sin(x)-cos(x).has.1.as.a.local.maximum.because.both.sin.and. cos.have.1.as.a.local.maximum. . 1 3.- .If.a.function.is.differentiable, then.it.must be.continuous. 1 4.> The.total-areas.of.step.function,.f(k)=k.feet, .k.ranging

1.>(40%) .True.and.False.1 1. = |sin(@)de = 0.9 - * 2 . f(x)=sin(x)-cos(x).has.1.as.a.local.maximum.because.both.sin.and. cos.have.1.as.a.local.maximum. . 1 3.- .If.a.function.is.differentiable, then.it.must be.continuous. 1 4.> The.total-areas.of.step.function,.f(k)=k.feet, .k.ranging from.1.to. 1000.is.500,500.ft2 . 1 If.f(x).is.continuous.on.a.closed.interval.I, and.f(a) and.f(b) .have. opposite.signs.where.a.and.bare.in.I, then.there.exists.a.value.c.in.[a,.b]. such.that.f.(c) .=0.1

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