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abclemons
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"Calculus" (Larson, et al) 9th ed: p. 169 #29: No match to answer key.
Locate the abs extrema on the interval of the function:
y=t-|t-3| for interval [-1,5]
|x|=[tex]\sqrt{x^{2}}[/tex]
I thought this would essentially be a subtraction rule and chain rule...
y'=1-((1/(2|t-3|))*2(t-3)*1)
y'=1-((t-3)/(|t-3|))
y'=(|t-3|-t+3)/|t-3|
Critical # at y=3
t(-1)=-5
t(3)=3
t(5)=3
abs maxima at (5,3) and (3,3)
abs minimum at (-1,-5)
Unfortunately, the answer key lists abs max (3,3) and abs min (-1,-1). I don't even get the (-1,-1) since t(-1) is -5...
If anyone has any guidance, please feel free to let it flow!
Homework Statement
Locate the abs extrema on the interval of the function:
y=t-|t-3| for interval [-1,5]
Homework Equations
|x|=[tex]\sqrt{x^{2}}[/tex]
The Attempt at a Solution
I thought this would essentially be a subtraction rule and chain rule...
y'=1-((1/(2|t-3|))*2(t-3)*1)
y'=1-((t-3)/(|t-3|))
y'=(|t-3|-t+3)/|t-3|
Critical # at y=3
t(-1)=-5
t(3)=3
t(5)=3
abs maxima at (5,3) and (3,3)
abs minimum at (-1,-5)
Unfortunately, the answer key lists abs max (3,3) and abs min (-1,-1). I don't even get the (-1,-1) since t(-1) is -5...
If anyone has any guidance, please feel free to let it flow!