Question

Find the approximate value of f(5.03), where f(x)=4x^2-7x+2.

(a) 67.99

(b) 56.99

(c) 67.66

(d) 78.09

This question was posed to me in an online quiz.

Question is taken from Derivatives Application topic in chapter Application of Derivatives of Mathematics – Class 12

Answer 1

Correct choice is (a) 67.99

To explain I would say: Let x=5 and Δx=0.03

Then, f(x+Δx)=4(x+Δx)^2-7(x+Δx)+2

Δy=f(x+Δx)-f(x)

∴f(x+Δx)=Δy+f(x)

Δy=f’ (x)Δx

⇒f(x+Δx)=f(x)+f’ (x)Δx

f(5.03)=(4(5)^2-7(5)+2)+(8(5)-7)(0.03) (∵f’ (x)=8x-7)

f(5.03)=(100-35+2)+(40-7)(0.03)

f(5.03)=67+33(0.03)

f(5.03)=67+0.99=67.99