r(t)=(t,t2) if t∈[−2,0]
=(t,t) if t∈(0,2)=(t,t2) if t∈[2,3]
(a) Domain of I is the set of all parts t Where I is defined. So, Domain of I=[−2,3]
(b)I(o)=(0,0)
limt→0−1I(t)=limt→0−1(t,t2)=(0,0)limt→0+1I(t)=limt→0+1(t,t2)=(0,0)
Since I(0)=limt→0I(t) , I is continuous at t=0.
(c)limt→2−1I(t)=limt→2−1(t,t)=(2,2)limt→2+1I(t)=limt→2+1(t,t2)=(2,4)
As Above both limits are not equal, So limt→2r(t) does npt exist. Hence r is not continuousat t=2.
(d)
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Let f be the function defined by f (x) = xe−x (a) Determine the y–intercept. (b) Determine the horizontal and vertical asymptotes. (c) Use the sign pattern for f'(x) to determine (i) the interval(s) over which f rises and where it falls; (ii) the local extrema. (d) Use the sign pattern for f ''(x) to determine (i) where the graph of f is concave up and where it is concave down (ii) the inflection point(s) (if any).
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