calculus for life science absolute max and min 9.10

Question

Find the absolute maximum and minimum values of each​ function, if they​ exist, over the indicated interval. Also indicate the​ x-value at which each extremum occurs.

​f(x)=8/3x^3−25/2x^2+3x+1​; (1,4)

A.The absolute maximum value is _____at x=_____, and the absolute minimum value is _____at x=____.

​(Use a comma to separate answers as needed. Type integers or decimals rounded to two decimal places as​ needed.)

B.The absolute minimum value is ____ at x=____ There is no absolute maximum.

​(Use a comma to separate answers as needed. Type integers or decimals rounded to two decimal places as​ needed.)

C.The absolute maximum value is ____ at x=___

There is no absolute minimum.

​(Use a comma to separate answers as needed. Type integers or decimals rounded to two decimal places as​ needed.)

Answer 1

8x^2 -25x + 3 = 0

(8x-1)(x-3) = 0

x=1, 3 = the x values for local extrema

check the endpoints, 1 and 4, for possible values for global extrema, as they may exceed the local, but not necessarily. but in this case, one is the same x value, so no difference. Local = global for 1, for the other end point 4 is close to 3.

f(x) =8x^3/3 - 25x^2/2 +3x + 1

take the derivative and set =0, then solve for x. then substitute those x values into f(x)

8-12.5+3+1 = -.5=y= local & global min, when x=1

8(4^3) -25(4^2)/2 +3(4) +1 = 512 -640/2+12+1=205= global max when x=4

8(3^3) -25(3^2)/2 +3(3) + 1 = 216 -112.5+9+1 = 226-112.5 = 113.5 = local max when x=3

-1/2 = local & global minimum (1,-.5)

113.5 = local maximum (3,113.5)

205 = global maximum (4,205)

or use a graphing calculator, on line or handheld

(no guarantees the above is error free)

205 = local & global maximum