Question

5.For \\(g(x)=\\frac{x-4}{x-3}, we can use the mean value theorem on [4, 6], Hence determine \\(c\\)
a.\\(\\sqrt (112) \\)
b.\\(c=2\\pm \\sqrt(3)}\\)
c.\\(c=-2\\pm \\sqrt(5)}\\)
d.\\(c=3\\pm \\sqrt(3)}\\)

6.Given\\(f(x)=\\sqrt(9-x^{2})\\)
a.\\(\\frac{-3}{8}\\)
b.\\(\\frac{5}{8}\\)
c.\\(\\frac{7}{8}\\)
d.\\(\\frac{-9}{8}\\)

Answer 1

QUESTION 5

Since the condition of the question says that we can use the mean value theorem on $[4,6]$, we will not prove it, but only use the formula

$$f'(c)=\frac{f(b)-f(a)}{b-a}$$

( More information: https://en.wikipedia.org/wiki/Mean_value_theorem )

In our case,

$$f(x)=\frac{x-4}{x-3}\rightarrow\begin{cases} f(6)=\displaystyle\frac{6-4}{6-3}=\frac{2}{3}\\[0.5cm] f(4)=\displaystyle\frac{4-4}{4-3}=0\\[0.5cm] f'(x)=\displaystyle\frac{(x-4)'(x-3)-(x-4)(x-3)'}{(x-3)^2}=\frac{1}{(x-3)^2} \end{cases}$$

Then,

$$f'(c)=\frac{f(6)-f(4)}{6-4}\rightarrow \frac{1}{(c-3)^2}=\frac{\displaystyle\frac{2}{3}-0}{2}\rightarrow (c-3)^2=3\\[0.5cm] |c-3|=\sqrt{3}\rightarrow c=3+\sqrt{3}\,\,\,{or}\,\,\,c=3-\sqrt{3}$$

Since $c\in[4,6]$ , then

$$\boxed{c=3+\sqrt{3}}$$

ANSWER

$$\boxed{d.\,\,\,c=3\pm\sqrt{3}}$$

Hint: We choose this answer, since only it contains the correct answer, although it also has an extra root.

QUESTION 6

Since the condition of the question is only $f(x)=\sqrt{9-x^2}$ , and there is no continuation, then I cannot answer this question.