If fx-x3-2x2-15x-36x-32- x - 3 p- x-3- is continuous at x-3- then the value of p is

Since f(x) is continuous at x=3, so we have f(3)=lim_x→ 3 f(x) ⇒ p=lim_x→ 3x^3 2x^2 15x+36(x 3)^2 =lim_x→ 3(x 3)^2(x+4)(x 3)^2 =lim_x→ 3 (x+4)=3+4=7 ...

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Standard XII

Mathematics

First Principle of Differentiation

If fx=x3-2x2-...

Question

If $f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{x^{3} - 2 x^{2} - 15 x + 36}{(x - 3)^{2}}, & x \neq 3 p, & x = 3 \end{matrix}$ is continuous at $x = 3$ , then the value of $p$ is

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Solution

Since $f (x)$ is continuous at $x = 3$ , so we have $f (3) = lim x \to 3 f (x)$
$\Rightarrow p = lim x \to 3 \frac{x^{3} - 2 x^{2} - 15 x + 36}{(x - 3)^{2}}$
$= lim x \to 3 \frac{(x - 3)^{2} (x + 4)}{(x - 3)^{2}}$
$= lim x \to 3 (x + 4) = 3 + 4 = 7$

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First Principle of Differentiation

Standard XII Mathematics

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If f(x)=⎧⎪⎨⎪⎩x3−2x2−15x+36(x−3)2,x≠3p,x=3 is continuous at x=3, then the value of p is

Since f(x) is continuous at x=3, so we have f(3)=limx→3f(x) ⇒p=limx→3x3−2x2−15x+36(x−3)2 =limx→3(x−3)2(x+4)(x−3)2 =limx→3(x+4)=3+4=7

If $f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{x^{3} - 2 x^{2} - 15 x + 36}{(x - 3)^{2}}, & x \neq 3 p, & x = 3 \end{matrix}$ is continuous at $x = 3$ , then the value of $p$ is

Since $f (x)$ is continuous at $x = 3$ , so we have $f (3) = lim x \to 3 f (x)$
$\Rightarrow p = lim x \to 3 \frac{x^{3} - 2 x^{2} - 15 x + 36}{(x - 3)^{2}}$
$= lim x \to 3 \frac{(x - 3)^{2} (x + 4)}{(x - 3)^{2}}$
$= lim x \to 3 (x + 4) = 3 + 4 = 7$