Limits, Mean Value Theorem, Intermediate Value Theorem, Piecewise Functions (40+ Problems)
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Solution Summary
Basic algebra problems are solved.
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as2pg1.jpg
1. Proof:
Forany e>0, we can find d=e/2>0, then for all x with |x-3|<d, we have
|f(x)-11|=|2x+5-11|=|2x-6|=2|x-3|<2d=2*e/2=e
Thus lim f(x) = 11 as x->3
2. Intermediate Value Theorem:
If f is continuous on a closed interval [a,b] and c is any number
between f(a) and f(b) inclusive, then there is at least one number x
in the closed interval such that f(x)=c.
as2pg2.jpg
3. Proof:
[f(x+t)-f(x)]/t
=[2(x+t)^2+3(x+t)-4-2x^2-3x+4]/t
=[4xt+3t+2t^2]/t
=4x+3+2t->4x+3
as t->0
Thus f'(x)=4x+3
4. Mean Value Theorem:
Let f be differentiable on the open interval (a,b) and continuous on the
closed interval [a,b], then there is at least one point c in (a,b), such
that f'(c)=[f(b)-f(a)]/(b-a).
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5. Proof:
Since 0<=|x^2 sin(1/x)|<=|x^2| and |x^2|->0 and 0->0 as x->0, thus
lim x^2 sin(1/x) = 0 as x->0
6. Suppose the shadow of the ball is x meters far from the hole. Let h be the
height of ...
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