Let f (x) be a polynomial of degree four having extreme values at x =1 an x =2. If then f(2) is equal to
-
-8
-
-4
-
0
-
4
C.
0
Any function have extreme values (maximum and minimum) at its critical points, where f'(x)= 0
Since, the function have extreme values at x =1 and x=2
therefore, f'(x) = 0 at x =1 and x= 2
⇒ f'(1) = 0 and f'(2) = 0
Also, it is given that
⇒ f(x) will be of the form
ax4 + bx3 + 2x4
f(x) is four degree polynomial]
Let f(x) = ax4 +bx3 +2x2
⇒ f'(x) = 4ax3 + 3bx2+ 4x
⇒ f'(1) = 4a +3b+4 = 0
and f'(2) 32a + 12b +8 = 0
⇒ 8a + 3b + 2 = 0
On solving Eqs. (i) and (ii), we get
a = 1/2, b = -2
f(2) = 8 - 16 +8 = 0