Question

The sum of all the local minimum values of the twice differentiable function f : R `rightarrow` R defined by

f(x) = `x^3 - 3x^2 - (3f^('')(2))/2 x + f^('')(1)`

Options

• – 22

• 5

• – 27

• 0

Answer 1

– 27

Explanation:

f(x) = `x^3 - 3x^2 - 3/2f^('')(2)x + f^('')(1)`

f'(x) = `3x^2 - 6x - 3/2f^('')(2)`

⇒ f''(x) = 6x – 6

Then, f'(2) = 12 – 6 = 6 and f'(1) = 0

∴ f(x) = x³ – 3x² – 9x

f'(x) = 3x² – 6x – 9 = 3(x + 1) (x – 3)

For critical points, f'(x) = 0

`\implies` 3x² – 6x – 9 = 0

`\implies` x = –1, 3

Now f'(x) = 6x – 6

f"(–1) = –12 < 0 `\implies` x = –1 is point of maxima

f"(3) = 12 > 0 `\implies` x = 3 is point of minima

So, local minimum value is, f(3) = 27 – 27 – 9 × 3 = – 27