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# |x|+x^2-5x+8

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f(x)=|x|+x^2-5x+8   define min and max for f(x) within [-1,4]

Guest Aug 5, 2015

#3
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Yeah, ty. I got a bit confused by the [-1,4] tried to find answers within that span instead of inserting [-1,4] as x.

Guest Aug 6, 2015
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#1
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in the interval [-1,0] the function is equal to

-x + x^2-5x+8 = x^2-6x+8

On this interval, this function has a minimum at x=0 (with y=8) and a max at x= -1 (with y=15)

On the interval [0,4] the function is equal to

x + x^2-5x+8 = x^2-4x+8

On this interval, this function has a minimum at x= -b/2a = 4/2 = 2 (with y=4) and a max at x=4 (with y=8)

over the entire interval from -1 to 4, the smallest value occurs when x=2 giving y=4, and the largest value occurs at x= -1 with y=15

Guest Aug 5, 2015
#2
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|x| splits at 0:

for x ≥ 0: |x| = x

for x < 0: |x| = -x

So:

for x ≥ 0: |x| + x2 - 5x + 8 becomes x + x2 - 5x + 8 --->   x2 - 4x + 8

for x  0: |x| + x2 - 5x + 8 becomes -x + x2 - 5x + 8 --->   x2 - 6x + 8

Both of these are (partial) parabolas.

The vertex of y = x2 - 4x + 8  occurs at (2,4) so the minimum value is 4.

The maximum value of x2 - 4x + 8 on [0,4] occurs at x = 4 --->  y = 8.

The maximum value of x2 - 6x + 8 on [-1,0] occurs at x = -1  --->   y = 15.

Thus, the maximum value is 15.

geno3141  Aug 5, 2015
#3
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