# The graph of h(x) is shown. The graph appears to be continuous at , where the definition changes. Show that h is in fact continuous at by finding the left and right limits and showing that the definition of continuity is met?

##  Feb 6, 2018

Kindly refer to the Explanation.

#### Explanation:

To show that $h$ is continuous, we need to check its

continuity at $x = 3$.

We know that, $h$ will be cont. at $x = 3$, if and only if,

${\lim}_{x \to 3 -} h \left(x\right) = h \left(3\right) = {\lim}_{x \to 3 +} h \left(x\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(\ast\right)$.

As $x \to 3 - , x < 3 \therefore h \left(x\right) = - {x}^{2} + 4 x + 1$.

$\therefore {\lim}_{x \to 3 -} h \left(x\right) = {\lim}_{x \to 3 -} - {x}^{2} + 4 x + 1 = - {\left(3\right)}^{2} + 4 \left(3\right) + 1$,

$\Rightarrow {\lim}_{x \to 3 -} h \left(x\right) = 4. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({\ast}^{1}\right)$.

Similarly, ${\lim}_{x \to 3 +} h \left(x\right) = {\lim}_{x \to 3 +} 4 {\left(0.6\right)}^{x - 3} = 4 {\left(0.6\right)}^{0}$.

$\Rightarrow {\lim}_{x \to 3 +} h \left(x\right) = 4. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({\ast}^{2}\right)$.

Finally, $h \left(3\right) = 4 {\left(0.6\right)}^{3 - 3} = 4. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({\ast}^{3}\right)$.

$\left(\ast\right) , \left({\ast}^{1}\right) , \left({\ast}^{2}\right) \mathmr{and} \left({\ast}^{3}\right) \Rightarrow h \text{ is cont. at } x = 3$.

Feb 6, 2018

See below:

#### Explanation:

For a function to be continuous at a point (call it 'c'), the following must be true:

• $f \left(c\right)$ must exist.

• ${\lim}_{x \to c} f \left(x\right)$ must exist

The former is defined to be true, but we'll need to verify the latter. How? Well, recall that for a limit to exist, the right and left hand limits must equal the same value. Mathematically:

${\lim}_{x \to {c}^{-}} f \left(x\right) = {\lim}_{x \to {c}^{+}} f \left(x\right)$

This is what we'll need to verify:

${\lim}_{x \to {3}^{-}} f \left(x\right) = {\lim}_{x \to {3}^{+}} f \left(x\right)$

To the left of $x = 3$, we can see that $f \left(x\right) = - {x}^{2} + 4 x + 1$. Also, to the right of (and at) $x = 3$, $f \left(x\right) = 4 \left({0.6}^{x - 3}\right)$. Using this:

${\lim}_{x \to 3} - {x}^{2} + 4 x + 1 = {\lim}_{x \to 3} 4 \left({0.6}^{x - 3}\right)$

Now, we just evaluate these limits, and check if they're equal:

$- \left({3}^{2}\right) + 4 \left(3\right) + 1 = 4 \left({0.6}^{3 - 3}\right)$

$\implies - 9 + 12 + 1 = 4 \left({0.6}^{0}\right)$

$\implies 4 = 4$

So, we have verified that $f \left(x\right)$ is continuous at $x = 3$.

Hope that helped :)