Let f(x)=(x^3-8)/(x-2), x≠ 2 12, x=2. (i) Show that f is continuous at x=2. (ii) Find f'(2).

Let $f (x) = {\frac{x ^{3} - 8}{x - 2}, 12, x \neq = 2 x = 2$ . (i) Show that $f$ is continuous at $x = 2$ . (ii) Find $f^{'} (2)$ .

hard 4 marks Class 12 · Mathematics · Continuity and Differentiability

Answer

The correct answer is 6 — Hence $f^{'} (2) = 6$ .

Factor the numerator: $x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4)$ , so for $x \neq = 2$ , $f (x) = x^{2} + 2 x + 4$ .
Compute the limit as $x \to 2$ : $2^{2} + 2 \cdot 2 + 4 = 12$ .
Since $f (2) = 12$ , the limit equals the function value, so $f$ is continuous at $2$ .
Differentiate for $x \neq = 2$ : $f^{'} (x) = 2 x + 2$ .
Take the limit of the derivative as $x \to 2$ : $2 \cdot 2 + 2 = 6$ .
Alternatively, use the definition: $lim_{h \to 0} \frac{f ( 2 + h ) - f ( 2 )}{h} = lim_{h \to 0} \frac{( 2 + h ) ^{2} + 2 ( 2 + h ) + 4 - 12}{h} = 6$ .
Hence $f^{'} (2) = 6$ .

Common mistake: Differentiating only the

x \neq = 2

part and ignoring the point of definition

Concept tested: Continuity and derivative of piecewise function via simplification