1 for some x ∊ (2 , 4) f " (x) = 1 for some x ∊ (2 , 4) f " (x) = 0 for some x ∊ (2 , 4) Solution: Consider f(x)=a x^{2}+b x+c f(2)=4 a+2 b+c=2 \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot(I) f(3)=9 a+3 b+c=5 \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot(\II) f(4)=16 a+4 b+c=10 \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot(\III) Subtract equation (I) from equation (II), 5 a+b=3 \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot(\IV) Subtract equation (II) from equation (III), 7 a+b=5 \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot(V) Subtract equation (IV) from equation (V), 2 a=2 a=1 Substitute the value in equation (IV), 5 \times 1+b=3 b=-2 Now, f^{'}(x)=2 x-2 f^{' '}(x)=2 since, f^{' '}(x)>1 for some x \in(2,4)" >
Consider f(x)=ax2+bx+c f(2)=4a+2b+c=2⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) f(3)=9a+3b+c=5⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II) f(4)=16a+4b+c=10⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(III) Subtract equation (I) from equation (II), 5a+b=3⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(IV) Subtract equation (II) from equation (III), 7a+b=5⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(V) Subtract equation (IV) from equation (V), 2a=2 a=1 Substitute the value in equation (IV), 5×1+b=3 b=−2 Now, f′(x)=2x−2 f′′(x)=2 since, f′′(x)>1 for some x∈(2,4)
