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Higher Order Derivatives

The function ...

Question

The function $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 8$ has maximum at x =

A

– 1

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B

2

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C

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D

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Solution

The correct option is A
– 1

$f^{'} (x) = 6 x^{2} - 6 x - 12$

f”(x) = 12x – 6

f’(x) = 0 $\Rightarrow$ $x^{2}$ – x – 2 = $\Rightarrow$ x = – 1, 2

f’’(-1) 0 $\Rightarrow$ f(x) has maxi. at x = – 1

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Similar questions

Q. Assertion (A): The function

f (x) = 2 x^{3} - 3 x^{2_{-}} 12 x + 8

has minimum value -12 at x = 2
Reason (R): For the fucntion

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 8, f^{'} (2) = 0

f^{''} (2) > 0

Q. The function

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4

Q. Let f(x) = 2x³

-

-

-

2, 4]. The relative maximum occurs at x=

(a)

-

-

1

(c) 2

(d) 4

Q. The function

y = 2 x^{3} - 3 x^{2} - 12 x + 8

has a local minimum at

x = ?

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}

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