Question

Find the approximate value of f(4.04), where f(x)=7x^3+6x^2-4x+3.

(a) 346.2

(b) 544.345

(c) 546.2

(d) 534.2

I had been asked this question in examination.

My question is from Derivatives Application topic in portion Application of Derivatives of Mathematics – Class 12

Answer 1

Correct choice is (c) 546.2

The explanation: Let x=4 and Δx=0.04

Then, f(x+Δx)=7(x+Δx)^3+7(x+Δx)^2-4(x+Δx)+3

Δy=f(x+Δx)-f(x)

∴f(x+Δx)=Δy+f(x)

Δy=f'(x)Δx

⇒f(x+Δx)=f(x)+f’ (x)Δx

Here, f'(x)=21x^2+12x-4

f(4.04)=(7(4)^3+6(4)^2-4(4)+3)+(21(4)^2+12(4)-4)(0.04)

f(4.04)=(448+96-16+3)+(336+48-4)(0.04)

f(4.04)=531+380(0.04)=546.2