(3) For what value of $$ a $$, does the function $$ f(x)=\left\{\begin{array}{ll}a x^{2}+1 & x \leq 2 \ 2 x-a & x2\end{array} ight. $$ have no points of discontinuity? $$ \quad a(2)^{2}+1=2(2)-a $$

Question

(3) For what value of $$ a $$, does the function $$ f(x)=\left\{\begin{array}{ll}a x^{2}+1 & x \leq 2 \ 2 x-a & x>2\end{array}ight. $$ have no points of discontinuity? $$ \quad a(2)^{2}+1=2(2)-a $$

UpStudy AI · Accepted Answer

To determine the value of $$ a $$ that ensures the function $$ f(x) $$ is continuous everywhere, we need to ensure that there is no discontinuity at $$ x = 2 $$. This involves making sure that the left-hand limit and the right-hand limit of $$ f(x) $$ at $$ x = 2 $$ are equal.

Given the function:
$$
f(x) = \begin{cases}
a x^{2} + 1 & \text{if } x \leq 2 \
2 x - a & \text{if } x > 2
\end{cases}
$$

1. **Left-Hand Limit as $$ x $$ approaches 2:**
   $$
   \lim_{x \to 2^-} f(x) = a(2)^2 + 1 = 4a + 1
   $$

2. **Right-Hand Limit as $$ x $$ approaches 2:**
   $$
   \lim_{x \to 2^+} f(x) = 2(2) - a = 4 - a
   $$

3. **Setting the Limits Equal for Continuity:**
   $$
   4a + 1 = 4 - a
   $$

4. **Solving for $$ a $$:**
   $$
   4a + a = 4 - 1 \
   5a = 3 \
   a = \frac{3}{5}
   $$

**Conclusion:**

For the function $$ f(x) $$ to have no points of discontinuity, the value of $$ a $$ must be $$ \frac{3}{5} $$.

$$
\boxed{\dfrac{3}{5}}
$$
$$ a = \frac{3}{5} $$

(3) For what value of , does the function have no points of
discontinuity?

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Algebra Questions

(3) For what value of , does the function have no points of discontinuity?