Question

1. Suppose that f(x) has a critical number at x=c, and f′′(c)=−10

By the Second Derivative Test, we conclude

• A. the test is inconclusive.
• B. x=c is an inflection point
• C. x=c is a local (relative) minimum
• D. x=c is a local (relative) maximum
• E. x=c is an absolute minimum

Question 4 of 10

3 Points

What follows is a numeric fill in the blank question with 2 blanks.

Find the absolute maximum and minimum value of the function

f(x)=0.5x^4+(4/3)x^3−3x^2+4

on the interval [0, 2]. Enter your answers with four decimal places in the following textboxes.

maximum = Blank 1. Fill in the blank, read surrounding text.
minimum = Blank 2. Fill in the blank, read surrounding text.

Answer 1

Ο 1Η χες is critica number of fx) Then by second derivative Test we koos that .- and too has locas mazuma at U=(, if f (o) <o ir fon has local minima aut nec, if f" (C) 70 f (0) = -10 <0 x=c, for bar loca marina. Therefore, the connect Ansajer & Inflection point is a point where f" () = 0 (c) = -10 70, Thus es an esflection point And here вре have, but 1=C not Answer is Therefore, the correct D. *EC is a local (relative) mancimom find the second Question: we have, have, f) = 0.5n+ +()x-374 fo is continuous [0,27 er find the critical (0,2) NOD we have to points of the on
The answer sheet has four pages.it is the first page
so. dift. f) with respect to x, De get d 8.5.4 t du 6.)»- 31'34 XB3 3x24 + 0 f'(x) = 0.5*4.4² + noun n-1 dx 22 + 4x² - 6x. For critical points of f(x) , we have Thus t'x)=0 gives 20+ 4212_640 oz, 270 ( +221-3) = 0 od 271 (x+3)(x-1)=0 By middle term factor Method " x=0, x=-3 and 1=1 Thun fa) has three critica point out no 1 X=-3 and x = 1 But only x=0,1 € [0, 2], this be have to consider only a=0 and [since 2= 0–3 & 10, 2] 2=1,
second page
Now, The greatest and lowest value of fo) on [0, 2] will give the absolute on [2] maruma and absolute absolete minima Now, fro) = 0+0 -0 + 4 = 4 f(1) = 0.5x (1) ² + 2 (1) ² - 3 (1) + 4 0.5 + 2/2 - 3+ 4 0.5 + + 3+ 4+' - 3+2+ 6 6 1.8333 2 1.8333 Corret epto four decine f(2)= 0.5%(2) + 4(2)3 3 (2) 2 9 = 0.516 + 4x8 places 1/2 x 8 - 3x4 + 4 8 + 32 3 - 12 + 4 32 3 10.66666 32 = 10.6667 Correct cepto four decional places
Third page
Thus we see that the lowest value of foo on [0, 2) is 1.8333 and et occurs at x = 1 Thus fon) has absolute minima at a=1 also see The greatest value of for) on and et occurs at 21=2 [0, 2] is 10.6667 absolete marima at a=2 Thus fa) has Therefore, maximum = 10.6667 N=2 misimum 108333 x = 1
Fourth/last page