Question

Find the approximate value of f (5.001), where f (x) = x3 − 7x2 + 15

Answer 1

Let x = 5 and ∆x = 0.001. Then, we have:

f(5.001)=f(x+∆x)=(x+∆x)=(x+∆x)3-7(x+∆x)2+15

Now,∆y=f(x+∆x)-f(x)

≈f(x)+f'(x).∆x (asdx≈∆x)

f(5.001)≈(x3-7x2+15)+(3x2-14x)∆x

-35+(5)(0.001)

-31+0.005

-34.995

Hence, the approximate value of f (5.001) is −34.995