Perform a first derivative test on the function f(x)-2x3 + 12x2-30x + 7; [-55].
ID: 2888275 • Letter: P
Question
Perform a first derivative test on the function f(x)-2x3 + 12x2-30x + 7; [-55]. a. Locate the critical points of the given function b. Use the first derivative test to locate the local maximum and minimum values. c. Identify the absolute minimum and maximum values of the function on the given interval (when they exist) a. Locate the critical points of the given function. Select the correct choice below and, if necessary, fill in the answer box within your choice. O A. The critical point(s) are located at x- O B. There are no critcal points. b. Locate the local maximum and minimum values. Select the correct choice below and, if necessary, fill in the answer box(es) within your choice. A. There are local maximum(s)at x; and local minimum(s) at x- O B. There are local minimum(s) atx-There are no local maximums. C. There are local maximum(s) at xThere are no local minimums. O D. There are no local maximums nor local minimums. c. ldentify the absolute minimum and maximum values of the function. Select the correct choice below and, if necessary, ill in the answer box(es) within your choice o A. The absolute maximum value is-and the absolute m (Use a comma to separate answers as needed.) (Use a comma to separate answers as needed.) (Use a comma to separate answers as needed.) (Use a comma to separate answers as needed.) minimum value is (Use a comma to separate answers as needed.) B. The absolute minimum value is . There is no absolute maximum. (Use a comma to separate answers as needed.)Explanation / Answer
given f(x) =2x3+12x2-30x+7
differentiate with respect to x
f '(x) =2*3x3-1+12*2x2-1 -30*1 +0
=> f '(x) =6x2+24x1 -30
=> f '(x) =6x2+24x -30
=> f '(x) =6(x2+4x -5)
=> f '(x) =6(x+5)(x-1)
a)
for critical points , f '(x) =0
=>6(x+5)(x-1)=0
=>x+5=0, x-1=0
=>x=-5,x=1
the critical points are located at x= -5 ,1
b)
at x =-5 , f '(x) changes from positive to negative
at x =1 , f '(x) changes from negative to positive
there are local maximum at x =-5 and local minimum at x= 1
c)
f(-5)= 207
f(1)= -9
f(5)=407
the absolute maximum value is 407, absolute minimum value is -9
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