Determine whether the following functions are defined and where they are continuous.

f(x)=\frac{x+\sin \pi x}{x^2+3x+2}, g(x)=\frac{x^2}{x+1}.

Solution

As \sin \pi x is defined everywhere, f(x) is defined as long as the denominator, x^2+3x+2\neq0. Now

x^2+3x+2=x^2+2x+x+2=x(x+2)+1(x+2)=(x+1)(x+2),

therefore x^2+3x+2\neq0\Leftrightarrow x\neq -1,-2. So f(x) is defined for all x\in\mathbb{R} except x=-1,-2. As x+\sin\pi x is continuous (as the sum of a polynomial and a sine) and x^2+3x+2 is continuous (as a polynomial), f(x) is continuous as long as x^2+3x+2\neq 0\Leftrightarrow x\neq-1,-2.

Similarly g(x) is defined for, and continuous at, all x+1\neq0\Leftrightarrow x\neq -1.

By evaluating f(0) and f(2), show that there is a solution to the equation 12(x+\sin\pi x)=x^2+3x+2.

Solution

f(0)=0 and f(2)=1/6. As f is continuous on the interval [0,2], f satisfies the hypothesis of the Intermediate Value Theorem on [0,2]:

As f is continuous, it takes all values between  0 and  1/6 — this is the Intermediate Value Theorem

Now, looking at the equation:

12(x+\sin\pi x)=x^2+3x+2

\Leftrightarrow f(x)= \frac{x+\sin\pi x}{x^2+3x+2}=\frac{1}{12},

for x\neq-1,-2 which is certainly the case here in [0,2]. Now 0\leq 1/12\leq 1/6 so by the Intermediate Value Theorem, there exists a c\in [0,2] such that f(c)=1/12 — which is equivalent to 12(x+\sin\pi x)=x^2+3x+2 having a solution.

Evaluate g(-2) and g(0), but show that there is no (real) solution to the equation g(x)=-2 lying in the interval (-2,0). What is different to your analysis of f(x)?

g(-2)=-4 and g(0)=0.

Let us find all solutions to g(x)=-2.

\frac{x^2}{x+1}=-2

\Rightarrow x^2=-2x-2,

which is fine as long as x\neq -1 (if x=-1, g(-1) is undefined so certainly not equal to -2.)

x^2+2x+2=0

\Rightarrow x_{\pm}=\frac{-2\pm\sqrt{4-4(2)}}{2}=\frac{-2\pm\sqrt{-4}}{2}\not\in\mathbb{R}.

That is there are no real solutions to g(x)=-2, in particular in the interval (-2,0).

Although g(-2)=-4 and g(0)=0, suggesting a c\in (-2,0) such that g(c)=-2, g does not in fact satisfy the hypothesis of the Intermediate Value Theorem on [-2,0] as it is not continuous at x=-1 and thus not continuous on [-2,0].

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