Question

7. Find both the maximum value and the minimum value of 3x4 - 8x3 + 12x2 - 48x + 25 on the interval [0, 3].

11th - 12th Class

Maths

Solution

194

4.0 (1 ratings)

Given, ( f(x)=3 x^{4}-8 x^{3}+12 x^{2}-48 x+25 ) differentiate with respect to ( mathrm{x} ), ( f^{prime}(x)=12 x^{3}-24 x^{2}+24 x-48 ) ( =12left(x^{3}-2 x^{2}+2 x-4right) ) ( =12left(x^{2}+2right)(x-2) ) now, ( f^{prime}(x)=0=12left(x^{2}+2right)(x-2) ) ( left(x^{2}+2right) eq 0 ) for all real value of ( x ) so, ( x=2 ) Now, we evaluate the value of ( f ) at critical point ( x=2 ) and at end points of the interval ( [0,3] . ) ( f(2)=3(2)^{4}-8(2)^{3}+12(2)^{2}-48(2)+25 ) ( =3(16)-8(8)+12(4)+25 ) ( =48-64+48-96+25 ) ( =-39 ) ( f(0)=3(0)^{4}-8(0)^{3}+12(0)^{2}-48(0)+25 ) ( =0+0+0+25 ) ( =25 ) ( f(3)=3(3)^{4}-8(3)^{3}+12(3)^{2}-48(3)+25 ) ( =3(81)-8(27)+12(9)+25 ) ( =243-216+108-144+25 ) ( =16 ) Therefore, we have the maximum value of ( f(x) ) on [0,3] is 25 occurring at ( x=0 ) And, the minimum value of ( f(x) ) on [0,3] is -39 occurring at ( x=2 ).