Let f(x) = 4+x, x<-4 9-x, x≥ -4 Find the indicated...

Let $\begin{aligned} f(x) = \begin{cases} 4+x, & x<-4\\ 9-x, & x\ge -4 \end{cases} \end{aligned}$ Find the indicated one-sided limits of $f$, and determine the continuity of $f$ at the indicated point. You should also sketch a graph of $y=f(x)$, including hollow and solid circles in the appropriate places. NOTE: Type DNE if a limit does not exist. $\lim_{x\to -4^-} f(x) =$ $\lim_{x\to -4^+} f(x) =$ $\lim_{x\to -4} f(x) =$ $f(-4) =$ Is $f$ continuous at $-4$? (YES/NO)

Transcript

00:01 In kappa t of this problem we are provided with the function f of x which equals to 1 minus x squared if x is less than or equal to 1 and it equals to natural log of x if x is greater than 1.

00:18 We are asked to show that the function f is continuous on the interval negative infinity up to infinity.

00:31 So for this, let us check if the function is continuous at each definition in the piecewise function.

00:41 So we have f of x to be equal to 1 minus x squared when x is less than or equal to 1.

00:48 So here, since we have a polynomial and we know that polynomials are continuous on negative infinity to infinity, it is clear that f of x equals to 1 minus x squared when x x.

01:01 Is less than or equal to 1 is continuous.

01:04 Next, we have f of x to be equal to natural log of x when x is greater than 1.

01:11 We also know that the logarithmic function is continuous for all positive values of x.

01:18 So it is clear that both the definitions of the function are continuous.

01:23 So all that is left is to check that if the function f of x is continuous, x equals to 1.

01:34 So for that let us first consider the left -hand limit.

01:38 So the left -hand limit is given by limit x tends to 1 -negative f -of -x.

01:45 By the definition of the function, we have limit x tends to 1 -negative of 1 -1 -x squared.

01:53 Substituting the value of x as 1, we get 1 -1 -1 which equals to 0.

01:58 So this is the left -hand limit.

02:01 Next, let us consider the right hand limiter.

02:06 So we have limit x tends to 1 positive of f of x and by the definition of the function we have limit x tends to 1 positive of natural log of x.

02:18 So substituting the value of x is 1 we have natural log of 1 which we know that equals to 0.

02:25 Now let us consider f of 1.

02:27 So we have f of 1 to be equal to 1 minus 1 squared which is 1 minus 1 and that equals to 0.

02:35 So since the left hand limit and the right hand limit are equal, we can say that limit x tends to 1 f of x equals to f of 1 which equals to 0.

02:47 So this implies that the function f is continuous on the interval negative infinity to infinity...

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