Question

If f(x) = `1/(4x^2 + 2x + 1); x ∈ R`, then find the maximum value of f(x).

Answer 1

f(x) = `1/(4x^2 + 2x + 1)`,

Let g(x) = 4x² + 2x + 1

= `4(x^2 + 2x 1/4 + 1/16) + 3/4`

= `4(x + 1/4)^2 + 3/4 ≥ 3/4`

∴ Maximum value of f(x) = `4/3`.

Answer 2

f(x) = `1/(4x^2 + 2x + 1)`,

Let f(x) = 4x² + 2x + 1

`\implies d/dx(g(x))` = g(x) = 8x + 2 and g(x) = 0 at x = `-1/4` also `d^2/(dx^2)(g(x))` = g"(x) = 8 > 0

`\implies` g(x) is minimum when x = `-1/4` so, f(x) is maximum at x = `-1/4`

∴ Maximum value of f(x) = `f(-1/4) = 1/(4(-1/4)^2 + 2(-1/4) + 1) = 4/3`.

Answer 3

f(x) = `1/(4x^2 + 2x + 1)`

On differentiating w.r.t x, we get

f'(x) = `(-(8x + 2))/(4x^2 + 2x + 1)^2`  ....(i)

For maxima or minima, we put

f'(x) = 0

`\implies` 8x + 2 = 0

`\implies` x = `-1/4`.

Again, differentiating equation (i) w.r.t. x, we get

f"(x) = `-{((4x^2 + 2x + 1)^2 (8) - (8x + 2)2 xx (4x^2 + 2x + 1)(8x + 2))/(4x^2 + 2x + 1)^4}`

At x = `-1/4, f^('')(-1/4) < 0`

f(x) is maximum at x = `-1/4`.

∴ Maximum value of f(x) is `f(-1/4) = 1/(4(-1/4)^2 + 2(-1/4) + 1) = 4/3`.

Answer 4

f(x) = `1/(4x^2 + 2x + 1)`

On differentiating w.r.t x,we get

f'(x) = `(-(8x + 2))/(4x^2 + 2x + 1)^2`  ....(i)

For maxima or minima, we put

f'(x) = 0

`\implies` 8x + 2 = 0

`\implies` x = `-1/4`.

When `x ∈ (-h - 1/4, -1/4)`, where 'h' is infinitesimally small positive quantity.

4x < – 1

`\implies` 8x < –2

`\implies` 8x + 2 < 0

`\implies` –(8x + 2) > 0 and (4x² + 2x + 1)² > 0

`\implies` f'(x) > 0

and when `x ∈ (-1/4, -1/4 + h)`, 4x > –1

`\implies` 8x > – 2

`\implies` 8x + 2 > 0

`\implies` – (8x + 2) < 0

and (4x² + 2x + 1)² > 0

`\implies` f'(x) < 0.

This shows that x = `-1/4` is the point of local maxima.

∴ Maximum value of f(x) is `f(-1/4) = 1/(4(-1/4)^2 + 2(-1/4) + 1) = 4/3`.