## "Calculus" (Larson, et al) 9th ed: p. 169 #29: No match to answer key.

1. The problem statement, all variables and given/known data
Locate the abs extrema on the interval of the function:
y=t-|t-3| for interval [-1,5]

2. Relevant equations
|x|=$$\sqrt{x^{2}}$$

3. 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!
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity

Mentor
 Quote by abclemons 1. The problem statement, all variables and given/known data Locate the abs extrema on the interval of the function: y=t-|t-3| for interval [-1,5] 2. Relevant equations |x|=$$\sqrt{x^{2}}$$ 3. 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!
It's much easier not to differentiate for this problem. On [-1, 3] the graph is a straight line with slope 2 and y-intercept -3. On [3, 5], the graph is horizontal.

I agree with you that the answer key is wrong. y(-1) = -5, not -1.
 typo? perhaps they want (1,-1) which is indeed a smaller abs(1-abs(1-3))=1 than abs(-1-abs(-1-3))=5 (edit: let me plot this to see...) edit: okay after plotting we have a line with positive slope 2, and a horizontal line intesecting at (3,3). so the critical points must be: (-1,-5), (3,3), (5,3) but in absolute value the first is the max and the second two are the min...sigh...

## "Calculus" (Larson, et al) 9th ed: p. 169 #29: No match to answer key.

@Mark44:
I agree that inspection of the graph is easier for this problem, but my professor likes for us to show critical numbers.

@xaos:
Once you pointed out the evaluation at 1 (rather than -1), a typo seems to be the most logical conclusion.

Respectfully to all,

Aaron

Recognitions:
Gold Member
Staff Emeritus
 Quote by abclemons @Mark44: I agree that inspection of the graph is easier for this problem, but my professor likes for us to show critical numbers.
Then show critical numbers- inspection of the graph shows that the derivative is 0 for all x greater than 3 and that the derivative does not exist at x= 3.

 @xaos: Once you pointed out the evaluation at 1 (rather than -1), a typo seems to be the most logical conclusion. Respectfully to all, Aaron

 Quote by HallsofIvy Then show critical numbers- inspection of the graph shows that the derivative is 0 for all x greater than 3 and that the derivative does not exist at x= 3.
I appreciate the input. I have set about working out analytical that for all real t>3 y'=0.

Respectfully,

Aaron

 Tags chain rule, critical number, extrema

 Similar discussions for: "Calculus" (Larson, et al) 9th ed: p. 169 #29: No match to answer key. Thread Forum Replies Biology, Chemistry & Other Homework 1 Biology, Chemistry & Other Homework 0 Calculus 3 Academic Guidance 7 General Discussion 106