Question:
For what value of \(k\),\[ f(x) =\left\{
\begin{array}{ll}
4x^2-kx-10 & 0 \leq x< 2\\
1-6x & 2\leq x \leq 3\\
\end{array}
\right.\] \(\text{is continuous} \quad \forall x\in [0,3]\).
Answer:
If the function is continuous, the limit at the break points must evaluate to the same value.
Therefore, the following condition must be met,
$$\lim_{x \to 2^-} 4x^2-kx-10=\lim_{x \to 2^+}1-6x$$
$$\Rightarrow 4(2)^2-k(2)-10=1-6(2)$$
$$\Rightarrow 6-2k=-11$$
$$\Rightarrow 17=2k$$
$$\Rightarrow k=\frac{17}{2}$$
Secondly, since the piecewise function consists of polynomials, we know that the polynomials are continuous for their respective domains. Hence, the value of k must be \(k=\frac{17}{2}\) if the function \(\text{is continuous} \quad \forall x\in [0,3]\).
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