For what value of the constant c is the function f continuous on the interval (−∞,∞).

solution:

f(x) has discontinuity at x=3

if it is continuous at x=3

then the continuous on (-\infty ,\infty )

we have,

f(x)= \left \{cx^{2}+8 ; if x< 3,x^{3}-cx ;if x\geq3

\lim_{x\rightarrow 3^-}f(x)=\lim_{x\rightarrow3^+}f(x)

\lim_{x\rightarrow3^-}(x^{2}+8x)=\lim_{x\rightarrow3^+}(x^{3}-cx)

(9c+24)= (227-3c)

9c+24=27-3c

9c+3c=27-24

12c=3

c=\frac{1}{4}

The mean value theorem states that, given a function continuous in some interval and differentiable in the same open interval , there exists a point such that

In your case, the function is a polynomial, and as such it is continuous and differentiable infinite times on the whole real number set. So, in particular, it is continuous in and differentiable in the same open interval .

Moreover, the derivative of the function is

So, the statement of the mean value theorem becomes the following: there exists a point such that

But we know that is constantly 1, so we have

which is an identity, i.e. an equation which is always verified.

So, every point in satisfies the mean value theorem.

Why? Well, the theorem states the existance of a point such that the tangent to that point (which is [/tex] f'(c) [/tex]) has the same slope as the secant connecting and .

But in this case, is a line, which means that the said secant and the function itself are actually the same thing.