Suppose that fx - x3 - 3x2 - 4x - 12 and hx - fx-x - 3 x - 3 K x - 3 - then find the value of K that makes -h- continuous at x - 3

f(x)=x^ 3 3x^ 2 4x+12 #160; #160; #160; #160; #160;h(x)= f(x) / x 3 x≠ 3 K x=3 f(x)=(x 2)(x+2)(x 3) # ...

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Properties of Conjugate of a Complex Number

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Question

Suppose that f(x) = $x^{3} - 3 x^{2} - 4 x + 12$ and h(x) = $⎧ ⎨ ⎩ \begin{matrix} \frac{f (x)}{x - 3} x \neq 3 K x = 3 \end{matrix}$ , then
find the value of K that makes 'h' continuous at x = 3

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

h (x) = [\begin{matrix} \frac{f (x)}{x - 3}, x \neq 3 K, x = 3 \end{matrix}

f

k

h

x = 3

k

h

Find the value of k, so that the function f(x) is continuous at the indicated point

$f (x) = \frac{\sqrt{3} - tan x}{π - 3 x} = k$ at

$x = \frac{π}{3}$

g (x) = lim n \to \infty \frac{x^{n} f (x) + h (x) + 1}{2 x^{n} + 3 x + 3},

x^{1} 1

g (1) = lim x \to 1 \frac{{sin}^{2} (π \cdot 2^{x})}{l n (sec (π \cdot 2^{x}))}

x = 1

4 g (1) + 2 f (1) - h (1)

f (x)

h (x)

x = 1

k

x = 0

f (x) = \frac{x^{3} - 3 x}{s i n x}

x \neq 0

= k

x = 0

Identify polynomials in the following :

(i) f(x) = $4 x^{3} - x^{2}$ - 3x + 7

(ii) g(x) = $2 x^{3} - 3 x^{2} + \sqrt{x} - 1$

(iii) p(x) = $\frac{2}{7} x^{2} - \frac{7}{4} x + 9$

(iv) q(x) = $2 x^{2} - 3 x + \frac{4}{x} + 2$

(v) h(x) = $x^{4} - x^{\frac{3}{2}} + x - 1$

(vi) f(x) = 2 + $\frac{3}{x} + 4 x$

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