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

[abclemons]

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

Homework Statement

Locate the abs extrema on the interval of the function:
y=t-|t-3| for interval [-1,5]

Homework Equations

|x|=\sqrt{x^{2}}

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!

 

[Mark44]

Homework Statement

Locate the abs extrema on the interval of the function:
y=t-|t-3| for interval [-1,5]

Homework Equations

|x|=\sqrt{x^{2}}

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.

 

[xaos]

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...

 

[abclemons]

@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

 

[HallsofIvy]

@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

 

[abclemons]

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