The intermediate value theorem states that suppose f is a function continuous on every point of the interval [a, b], for any L between the values f(a) and f(b), there exist at least one number, c, in [a, b] for which f(c) = L.
83.
Given
[tex]f(x)=x^2+x-1[/tex]
on the interval [0, 5], with f(c) = 11
[tex]f(0)=0^2+0-1=-1[/tex]
and
[tex]f(5)=5^2+5-1=25+4=29[/tex]
Clearly, f(c) = 11 is between f(0) = -1 and f(5) = 29
Thus, the intermediate value theorem is satisfied.
To find the value of c, we solve the equation
[tex]c^2+c-1=11[/tex]
as follows:
[tex]c^2+c-1=11 \\ \\ c^2+c-12=0 \\ \\ (c-3)(c+4)=0 \\ \\ c=3 \ or \ -4[/tex]
Because, c is in the interval [0, 5], therefore, the value of c is 3.
84.
Given
[tex]f(x)=x^2-6x+8[/tex]
on the interval [0, 3], with f(c) = 0
[tex]f(0)=0^2-6(0)+8=8[/tex]
and
[tex]f(3)=3^2-6(3)+8=9-18+8=-1[/tex]
Clearly, f(c) = 0 is between f(0) = 8 and f(3) = -1
Thus, the intermediate value theorem is satisfied.
To find the value of c, we solve the equation
[tex]c^2-6c+8=0[/tex]
as follows:
[tex]c^2-6c+8=0 \\ \\ (c-4)(c-2)=0 \\ \\ c=4 \ or \ 2[/tex]
Because, c is in the interval [0, 3], therefore, the value of c is 2.
85.
Given
[tex]f(x)=x^3-x^2+x-2[/tex]
on the interval [0, 3], with f(c) = 4
[tex]f(0)=0^3-0^2+0-2=-2[/tex]
and
[tex]f(3)=3^3-3^2+3-2=27-9+1=19[/tex]
Clearly, f(c) = 4 is between f(0) = -2 and f(3) = 19
Thus, the intermediate value theorem is satisfied.
To find the value of c, we solve the equation
[tex]c^3-c^2+c-2=4[/tex]
as follows:
[tex]c^3-c^2+c-6=0 \\ \\ (c^2+c+3)(c-2)=0 \\ \\ c=2[/tex]
Therefore, the value of c is 2.
86.
Given
[tex]f(x)= \frac{x^2+x}{x-1} [/tex]
on the interval [tex]\left[ \frac{5}{2}, \ 4\right] [/tex], with f(c) = 6
[tex]f\left(\frac{5}{2}\right)= \frac{\left(\frac{5}{2}\right)^2+\left(\frac{5}{2}\right)}{\left(\frac{5}{2}\right)-1}= \frac{ \frac{25}{4}+ \frac{5}{2} }{ \frac{3}{2} } = \frac{35}{4} \cdot \frac{2}{3} =5 \frac{5}{6} [/tex]
and
[tex]f(4)= \frac{4^2+4}{4-1}= \frac{ 16+ 4 }{ 3 } = \frac{20}{3}=6 \frac{2}{3}[/tex]
Clearly, f(c) = 6 is between [tex]f\left(\frac{5}{2}\right)= 5 \frac{5}{6} [/tex] and [tex]f(4)=6 \frac{2}{3}[/tex]
Thus, the intermediate value theorem is satisfied.
To find the value of c, we solve the equation
[tex]\frac{c^2+c}{c-1}=6[/tex]
as follows:
[tex]c^2+c=6(c-1)=6c-6 \\ \\ \Rightarrow c^2-5c+6=0 \\ \\ \Rightarrow(c-2)(c-3)=0 \\ \\ \Rightarrow c=2 \ or \ 3[/tex]
Because, c is in the interval [tex]\left[ \frac{5}{2}, \ 4\right] [/tex], therefore, the value of c is 3.
