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Stationary Point

The stationar...

Question

The stationary points of $f (x) = x^{3} - 9 x^{2} + 24 x - 12$ are

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Solution

$f^{'} (x) = 3 x^{2} - 18 x + 24 = 3 (x^{2} - x + 8) = 3 (x - 4) (x - 2) f o r s t a t i o n a r y p o i n t s f^{'} (x) = 0 ∴ x = 2, 4$

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Similar questions

Q. Examine following function for maximum and minimum

f (x) = x^{3} - 9 x^{2} + 24 x

Q. The maximum value of

f (x) = x^{3} - 9 x^{2} + 24 x + 5

in the interval

[1, 6]

Q. Find the roots of

x^{3} - 9 x^{2} + 24 x - 16 = 0

Q. Find the intervals in which the following functions are increasing or decreasing.
(i) f(x) = 10 − 6x − 2x²

(ii) f(x) = x² + 2x − 5

(iii) f(x) = 6 − 9x − x²

(iv) f(x) = 2x³ − 12x² + 18x + 15

(v) f(x) = 5 + 36x + 3x² − 2x³

(vi) f(x) = 8 + 36x + 3x² − 2x³

(vii) f(x) = 5x³ − 15x² − 120x + 3

(viii) f(x) = x³ − 6x² − 36x + 2

(ix) f(x) = 2x³ − 15x² + 36x + 1

(x) f(x) = 2x³ + 9x² + 12x + 20

(xi) f(x) = 2x³ − 9x² + 12x − 5

(xii) f(x) = 6 + 12x + 3x² − 2x³

(xiii) f(x) = 2x³ − 24x + 107

(xiv) f(x) = −2x³ − 9x² − 12x + 1

(xv) f(x) = (x − 1) (x − 2)²

(xvi) f(x) = x³ − 12x² + 36x + 17

(xvii) f(x) = 2x³ − 24x + 7

(xviii)

f (x) = \frac{3}{10} x^{4} - \frac{4}{5} x^{3} - 3 x^{2} + \frac{36}{5} x + 11

(xix) f(x) = x⁴ − 4x

(xx)

f (x) = \frac{x^{4}}{4} + \frac{2}{3} x^{3} - \frac{5}{2} x^{2} - 6 x + 7

(xxi) f(x) = x⁴ − 4x³ + 4x² + 15

(xxii) f(x) = 5x³^/2 − 3x⁵^/2, x > 0

(xxiii) f(x) = x⁸ + 6x²

(xxiv) f(x) = x³ − 6x² + 9x + 15

(xxv)

f (x) = {\{x (x - 2)\}}^{2}

Q. I: The function

f (x) = 9 x^{2} - 15 x - x^{3} + 10

is increasing in

(1, 5)

II: The function

f (x) = 9 x^{2} - 15 x - x^{3} + 10

is decresing in

(1, 5)

Which of the above statements are true ?

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