Calculus - 11

+2 votes

Find maximum value of f(x)= 2x^3-15x^2+36x+10

(A) 37

(B) 27

(C) 38

(D) 42

asked May 30 in Basic Maths by gbmentor (14,290 points)
reshown May 31 by gbmentor

4 Answers

+1 vote
 
Best answer

differentiate\ f(x)\ and\ equate\ it\ to\ 0 to\ find\ critical\ points\ \\ f'(x)= 6x^2-30x+36=0\\ f'(x)=x^2-5x+6=0\\ f'(x)=x^2-2x-3x+6=0\\ f'(x)=(x-2)(x-3)=0\\ \therefore x=2,3\\ Again\ differentiate\ f'(x)\\ f''(x)=12x-30\\ put\ x=2\ in\ f''(x)\\ f''(2)= 24-30=-6<0 \ \therefore Maximum\\ put\ x=3\ in\ f''(x)\\ f''(3)=36-30=+6>0\ \therefore Minimum\\ So\ maximum\ value\ of\ f(x)\ at\ x=2\\ f(2)=2*8-15*4+36*2+10\\ f(2)=16-60+72+10\\ f(2)= 38

Therefore ans must be C.

answered May 31 by tskushagra-guptacse (10,800 points)
selected May 31 by getgatebook
0 votes
38

 

Find df/dx equate to zero

U get critical points....x1

Find 2nd derivative of x...and substitute critical points....

If <0 maximum

Find f(x1) which is maximum value
answered May 31 by harish29pro20 (640 points)
0 votes
question seems to be wrong.

if it is asking maximum value then it should give range too.

otherwise it should be local maxima value
answered May 31 by tskaran25gupta (770 points)
Yes...we need to take local maximum value
0 votes
wrong question , 38 would have been the answer if it was local maxima ...

Sir there are a lot of wrong questions coming up plzz.... chk
answered May 31 by tssoumambanerjee-nit (490 points)
Answer:
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