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Polynomial

If fx = 2x3...

Question

If $f (x) = 2 x^{3} + 3 x^{2} - 11 x + 6$ , find $f (- 1)$

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Solution

In the given polynomial $f (x) = 2 x^{3} + 3 x^{2} - 11 x + 6$ , substitute $x = - 1$ as shown below:

$f (- 1) = (2 \times (- 1)^{3}) + (3 \times (- 1)^{2}) - (11 \times (- 1)) + 6 = (2 \times - 1) + (3 \times 1) + 11 + 6 = - 2 + 3 + 11 + 6 = - 2 + 20$
$= 18$

Hence, $f (- 1) = 18$ .

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f (x) = 2 x^{3} + 3 x^{2} - 11 x + 6

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f (x) = 2 x^{3} + 3 x^{2} - 11 x + 6

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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

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Verify whether -2, $\frac{1}{2}$ are zeroes of $2 x^{3} - 3 x^{2} - 11 x + 6$ . [2 MARKS]

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