Answer:
There is a value of [tex]c[/tex] in (-1, 1), [tex]c = 0.577[/tex].
Step-by-step explanation:
Let [tex]f(x) = x^{3}+2\cdot x[/tex] for [tex]x \in[-1,1][/tex], we need to prove that [tex]f(x)[/tex] is continuous and differentiable to apply the Mean Value Theorem. Given that [tex]f(x)[/tex] is a polynomical function, its domain comprises all real numbers and therefore, function is continuous.
If [tex]f(x)[/tex] is differentiable, then [tex]f'(x)[/tex] exists for all value of [tex]x[/tex]. By definition of derivative, we obtain the following expression:
[tex]f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]
[tex]f'(x) = \lim_{h \to 0} \frac{(x+h)^{3}+2\cdot (x+h)-x^{3}-2\cdot x}{h}[/tex]
[tex]f'(x) = \lim_{h \to 0} \frac{x^{3}+3\cdot x^{2}\cdot h+3\cdot x\cdot h^{2}+h^{3}+2\cdot x+2\cdot h-x^{3}-2\cdot x}{h}[/tex]
[tex]f'(x) = \lim_{h \to 0} \frac{3\cdot x^{2}\cdot h+3\cdot x\cdot h^{2}+h^{3}+2\cdot h}{h}[/tex]
[tex]f'(x) = \lim_{h \to 0} 3\cdot x^{2}+ \lim_{h \to 0} 3\cdot x \cdot h+ \lim_{h \to 0} h^{2}+ \lim_{h \to 0} 2[/tex]
[tex]f'(x) = 3\cdot x^{2}+2[/tex] (Eq. 2)
The derivative of a cubic function is quadratic function, which is also a polynomic function. Hence, the function is differentiable at the given interval.
According to the Mean Value Theorem, the following relationship is fulfilled:
[tex]f'(c) = \frac{f(1)-f(-1)}{1-(-1)}[/tex] (Eq. 3)
If we know that [tex]f(-1) = -3[/tex], [tex]f(1) = 3[/tex] and [tex]f'(c) = 3\cdot c^{2}+2[/tex], then we expand the definition as follows:
[tex]3\cdot c^{2}+2 = 3[/tex]
[tex]3\cdot c^{2} = 1[/tex]
[tex]c = \sqrt{\frac{1}{3} }[/tex]
[tex]c \approx 0.577[/tex]
There is a value of [tex]c[/tex] in the interval (-1, 1), [tex]c = 0.577[/tex].
