Question

Find the approximate value of f (5.001), where f (x) = x^ 3 − 7x^ 2 + 15.

Answer 1

HELLO DEAR,

GIVEN:- f(x) = x³ - 7x² + 15.

f'(x) = 3x² -14x.

we know, f(x + ∆x) = f'(x) * ∆x

we may write, 5.001 = (5 + 0.001)
put x = 5 , ∆x = 0.001,

f(5 + 0.001) - f(5) = (3x² - 14x)* (0.001)

since, x = 5

f(5.001) = (5)³ - 7(5)² + 15 + (3*5² - 14*5)(0.001)

f(5.001) = 125 - 172 + 15 + (0.001)(75 - 70)

f(5.001) = -32 + 0.005 = -34.995

hence, the approximate value of f(5.001) = -34.995


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

Step-by-step explanation:

Hope you understood.