Answer:
See below in bold.
Step-by-step explanation:
The x intercepts occur when f(x) = 0.
1. logx - 1 = 0
logx = 1
By the definition of a log ( to the base 10):
x = 10^1 = 10
So the x-intercept is c (10,0).
2. - (logx - 2) = 0
logx - 2 = 0
log x = 2
so x = 100.
So it is d (100,0).
3 . log(-x - 2) = 0
-x - 2 = 10^0 = 1
-x = 3
x = -3
So it is b (-3, 0).
4. f(x) = -log -(x - 1)
log - (x - 1) = 0
log 1 = 0
so -(x - 1) = 1
- x + 1 = 1
x = 1-1 = 0
So it is a. (0,0).
Function x-intercept
[tex]f(x)=\log x-1[/tex] [tex](10,0)[/tex]
[tex]f(x)=-(\log x-2)[/tex] [tex](100,0)[/tex]
[tex]f(x)=\log (-x-2)[/tex] [tex](-3,0)[/tex]
[tex]f(x)=-\log -(x-1)[/tex] [tex](0,0)[/tex]
We know that the x-intercept of a function is the point where the function value is zero.
i.e. the x where f(x)=0
1)
[tex]f(x)=\log x-1[/tex]
when [tex]f(x)=0[/tex] we have:
[tex]\log x-1=0\\\\i.e.\\\\\log x=1\\\\i.e.\\\\\log x=\log 10[/tex]
Hence, taking the exponential function on both the sides of the equation we have:
[tex]x=10[/tex]
The x-intercept is: (10,0)
2)
[tex]f(x)=-(\log x-2)[/tex]
when, [tex]f(x)=0[/tex]
we have:
[tex]-(\log x-2)=0\\\\i.e.\\\\\log x-2=0\\\\i.e.\\\\\log x=2\\\\i.e.\\\\\log x=2\cdot 1\\\\i.e.\\\\\log x=2\cdot \log 10\\\\i.e.\\\\\log x=\log (10)^2[/tex]
Since,
[tex]m\log n=\log n^m[/tex]
Hence, we have:
[tex]\log x=\log 100[/tex]
Taking anti logarithm on both side we get:
[tex]x=100[/tex]
Hence, the x-intercept is:
(100,0)
3)
[tex]f(x)=\log (-x-2)[/tex]
when
[tex]f(x)=0[/tex]
we have:
[tex]\log (-x-2)=0\\\\i.e.\\\\\log (-x-2)=\log 1[/tex]
On taking anti logarithm on both the side of the equation we get:
[tex]-x-2=1\\\\i.e.\\\\x=-2-1\\\\i.e.\\\\x=-3[/tex]
Hence, the x-intercept is: (-3,0)
4)
[tex]f(x)=-\log -(x-1)[/tex]
when,
[tex]f(x)=0\ we\ have:[/tex]
[tex]-\log -(x-1)=0\\\\i.e.\\\\\log -(x-1)=0\\\\i.e.\\\\\log -(x-1)=\log 1\\\\i.e.\\\\-(x-1)=1\\\\i.e.\\\\x-1=-1\\\\i.e.\\\\x=-1+1\\\\i.e.\\\\x=0[/tex]
Hence, the x-intercept is: (0,0)
