Q3. Pour quelle valeur du paramètre a la fonction suivante est-elle continue? $f(x) = \begin{cases} \frac{x^2 - 4x - 32}{x - 8} & \text{si } x \neq 8 \\ a & \text{si } x = 8 \end{cases}$
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Step 1: For the function to be continuous at $x = 8$, the limit of the function as $x$ approaches 8 must be equal to the value of the function at $x = 8$, which is $a$. Show more…
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