Question
Mathematics Question on Maxima and Minima
Find the approximate value of f(3.01), where f(x)=3x2+3.
A
(A) 30.18
B
(B) 30.018
C
(C) 30.28
D
(D) 30.08
Answer
(A) 30.18
Explanation
Solution
Explanation:
Let, small charge in x be Δx and the corresponding change in y is Δy.Δy=dydxΔx=f′(x)ΔxNow that Δy=f(x+Δx)−f(x)Therefore, f(x+Δx)=f(x)+ΔyGiven: f(x)=3x2+3Let, x+Δx=3.01=3+0.01Therefore, x=3 and Δx=0.01f(x+Δx)=f(x)+Δy⇒f(x+Δx)=f(x)+f′(x)Δx⇒f(3.01)=3x2+3+(6x)Δx⇒f(3.01)=3(3)2+3+(6⋅3)(0.01)⇒f(3.01)=30+0.18⇒f(3.01)=30.18Hence, the correct option is (A).