Find the absolute extrema of the function (if any exist) on each interval. (If an answer does not exist, enter DNE.)

Question

f(x) = x² – 6x

a) (-1,6)

Min: 3,-9

Max: I got it wrong as DNE and -1,5. Answer ?

b) (0,6)

Min: got wrong as 6,0 and 0,0. Answer ?

Max: DNE

Answer 1

In an up-wad opening parabola the minimum exists at x = -b /2a.

I this case x = -(-6)/2(1) or x = 3.

3 is with within the given interval.

The minimum is (3)² - 6(3), or -9

Answer 2

This comes under the section of the EVT: extreme value theorem. There shouldn't be any DNE answers there's no discontinuities.

ƒ(x) = x² - 6x ƒ´(x) = y´ = 2x - 6

a) Interval [-1,6] *Note: I wondering if these are brackets, rather than having to approach those values.

Take you interval values first.

ƒ(-1) = 1 + 6 = 7

ƒ(6) = 36 - 36 = 0

Your your derivative to zero. When m = 0 that indicates that the original may have hit a top or bottom.

y´ = 2x - 6 = 0 ∴ x = 3, now plug 3 back into the original equation.

ƒ(3) = 9 - 18 = -9

∴ (therefore) our max and min occur at WHERE? when x = -1 and x = 3

So, max at (-1, 7) and min at (3, -9)

b) Interval of [0, 6]

So, take the values.

ƒ(0) = 0

ƒ(6) = 36 - 36 = 0

y´ = 2x - 6 = 0 ∴ x = 3, same procedure as above. Plug 3 back into the original equation.

ƒ(3) = 9 - 18 = -9

∴ our absolute max and absolute min are (0, 0), (6,0) and (3, -9)