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प्रश्न
Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has
- local maxima
- local minima
- point of inflexion
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उत्तर
Here, f (x) = (x - 2)4 (x + 1)4
∴ f'(x) = (x – 2)4 · 3(x + 1)2 + (x + 1)3 · 4(x - 2)3
= (x – 2)3 (x + 1)2 [3(x – 2) + 4(x + 1)]
= (x – 2)3 (x + 1)2 [3x – 6 + 4x + 4]
= (x – 2)3 (x + 1)2 (7x – 2)
= 7(x - 2)3 (x + 1)2 `(x - 2/7)`
For maximum/minimum, 1(x) = 0
⇒ 7(x - 2)3 + (x + 1)2 `(x - 2/7)` = 0
∴ x = 2, -1, `2/7`
(i) When x = 2,
x is near 2 and to the left of 2 then, f(x) = (-)(+)(+) = -ve
x is near 2 and to the right of 2 then, f(x) = (+)(+)(+) = + ve
∴ The sign of f(x) changes from negative to positive as x passes through x = -2.
⇒ f is minimum at x = 2.
(ii) At, x = -1
For values of x near -1 and less than 1,
f'(x) = (-)(+)(-) = + ve
For values of x near to -1 and greater than -1,
f(x) = (-)(+)(-) = + ve
does not change its sign while passing through the point x = -1.
⇒ Thus, x = -1 is a point of inflexion
(iii) At, x = `2/7` = 0.28
On placing the value of x less than `2/7` near `2/7`,
f'(x) = (-)(+)(-) = + ve
Keeping the value of x close to `2/7` and greater than `2/7`,
f'(x) = (-)(+)(-) = -ve
⇒ At x = `2/7`, (x) changes from positive to negative.
As x passes through, x = `2/7`.
Thus it is minimum at x = 2, inflexion at x = -1 and a maximum at x = `2/7`.
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