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?

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2 Answers
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(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.

Feb 6, 2018

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 :)