Question

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?

Answer 1

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(x)=h(3)=lim_(x to 3+) h(x)...............................(ast)`.

As `x to 3-, x lt 3 :. h(x)=-x^2+4x+1`.

`:. lim_(x to 3-)h(x)=lim_(x to 3-)-x^2+4x+1=-(3)^2+4(3)+1`,

`rArr lim_(x to 3-)h(x)=4......................................................(ast^1)`.

Similarly, `lim_(x to 3+)h(x)=lim_(x to 3+)4(0.6)^(x-3)=4(0.6)^0`.

`rArr lim_(x to 3+)h(x)=4....................................................(ast^2)`.

Finally, `h(3)=4(0.6)^(3-3)=4......................................(ast^3)`.

`(ast),(ast^1),(ast^2) and (ast^3) rArr h" is cont. at "x=3`.

Answer 2

See below:

Explanation:

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

• `f(c)` must exist.

• `lim_(x->c)f(x)` 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->c^-)f(x) = lim_(x->c^+)f(x)`

This is what we'll need to verify:

`lim_(x->3^-)f(x) = lim_(x->3^+)f(x)`

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

`lim_(x->3)-x^2 +4x + 1 = lim_(x->3)4(0.6^(x-3))`

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

`-(3^2) + 4(3) + 1 = 4(0.6^(3-3))`

`=> -9 + 12 + 1 = 4(0.6^0)`

`=> 4 = 4`

So, we have verified that `f(x)` is continuous at `x = 3`.

Hope that helped :)