Question

For the function f(x) = 4x³ + 3x² – 6x + 1 find the intervals of monotonicity, local extrema, intervals of concavity and points of inflection

Answer 1

(x) = 4x³ + 3x² – 6x + 1

Monotonicity

f(x) = 4x³ + 3x² – 6x + 1

f'(x) = 12x² + 6x – 6

f'(x) = 0

⇒ 6(2x² + x – 1) = 0

x = – 1, `1/2`   ......(Stationary points)

∴ The intervals of monotonicity are `(- oo, -1)(-1, 1/2)` and `(1/2, oo)`

In `(-oo, -1)`, f'(x) > 0 ⇒ f(x) is strictly increasing

In `(- 1, 1/2)`, f'(x) < 0 ⇒ f(x) is strictly decreasing

In `(1/2, 00)`, f'(x) > 0 ⇒ f(x) is strictly increasing

f(x) attains local maximum as f'(x) changes its sign from positive to negative when passing through x = – 1

∴ Local maximum f(– 1) = – 4 + 3 + 6 + 1 = 6

f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = `1/2`

∴ Local minimum `"f"(1/2)`

= `4(1/8) + 3(1/4) - 6(1/2) + 1`

= `1/2 + 3/4 - + 1 = - 3/4`

f(x) = 4x³ + 3x² – 6x + 1

f'(x) = 12x² + 6x – 6

f”(x) = 24x + 6

f’(x) = 0

⇒ 24x + 6 = 0

x = `- 6/24 = - 1/4`  ......(Critical points)

∴ The intervals are `(oo, 1/4)` and `(1/4, oo)`, f”(x) > 0

In the interval `(- oo, - 1/4)`, f”(x) < 0 ⇒ curve is concave down.

In the interval `(- 1/4, 00)`, f”(x) > 0 ⇒ curve is concave up.

The curve is concave upward in `(- 1/4, oo)` and concave downward in `(- oo, - 1/4)`

f”(x) changes its sign when passing through x = `-1/4`

Now, `"f"(- 1/4) = 4(- 1/64) + 3(1/16) - 6(- 1/4) + 1`

= `1/16 + 3/16 + 3/2 + 1`

= `21/8`

∴ The point of inflection is `(-1/4, 21/8)`