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