Answer :
Sure! Let's go through the problem step by step.
### Part a
We need to find the output for [tex]\( f(x) = x^2 - 1 \)[/tex] when the input is [tex]\( x = 4 \)[/tex].
To do this, we substitute [tex]\( x = 4 \)[/tex] into the function:
[tex]\[ f(4) = 4^2 - 1 = 16 - 1 = 15 \][/tex]
So, the output when the input is [tex]\( x = 4 \)[/tex] is:
[tex]\[ f(4) = 15 \][/tex]
### Part b
Next, we need to find [tex]\( f(-1) \)[/tex] and [tex]\( f(10) \)[/tex].
#### For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^2 - 1 = 1 - 1 = 0 \][/tex]
So, the output when the input is [tex]\( x = -1 \)[/tex] is:
[tex]\[ f(-1) = 0 \][/tex]
#### For [tex]\( x = 10 \)[/tex]:
[tex]\[ f(10) = 10^2 - 1 = 100 - 1 = 99 \][/tex]
So, the output when the input is [tex]\( x = 10 \)[/tex] is:
[tex]\[ f(10) = 99 \][/tex]
### Part c
We need to find the input [tex]\( x \)[/tex] when the output of the function is 24. That is, we need to solve for [tex]\( x \)[/tex] in the equation [tex]\( f(x) = 24 \)[/tex].
The function is:
[tex]\[ f(x) = x^2 - 1 \][/tex]
If [tex]\( f(x) = 24 \)[/tex]:
[tex]\[ x^2 - 1 = 24 \][/tex]
[tex]\[ x^2 = 24 + 1 \][/tex]
[tex]\[ x^2 = 25 \][/tex]
By taking the square root of both sides, we get:
[tex]\[ x = \pm 5 \][/tex]
So, there are two possible inputs for which the output is 24:
[tex]\[ x = 5 \][/tex] or [tex]\( x = -5 \)[/tex]
### Summary
- The output for [tex]\( f(4) \)[/tex] is [tex]\( 15 \)[/tex].
- The output for [tex]\( f(-1) \)[/tex] is [tex]\( 0 \)[/tex].
- The output for [tex]\( f(10) \)[/tex] is [tex]\( 99 \)[/tex].
- The inputs for which [tex]\( f(x) = 24 \)[/tex] are [tex]\( x = 5 \)[/tex] and [tex]\( x = -5 \)[/tex].
### Part a
We need to find the output for [tex]\( f(x) = x^2 - 1 \)[/tex] when the input is [tex]\( x = 4 \)[/tex].
To do this, we substitute [tex]\( x = 4 \)[/tex] into the function:
[tex]\[ f(4) = 4^2 - 1 = 16 - 1 = 15 \][/tex]
So, the output when the input is [tex]\( x = 4 \)[/tex] is:
[tex]\[ f(4) = 15 \][/tex]
### Part b
Next, we need to find [tex]\( f(-1) \)[/tex] and [tex]\( f(10) \)[/tex].
#### For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^2 - 1 = 1 - 1 = 0 \][/tex]
So, the output when the input is [tex]\( x = -1 \)[/tex] is:
[tex]\[ f(-1) = 0 \][/tex]
#### For [tex]\( x = 10 \)[/tex]:
[tex]\[ f(10) = 10^2 - 1 = 100 - 1 = 99 \][/tex]
So, the output when the input is [tex]\( x = 10 \)[/tex] is:
[tex]\[ f(10) = 99 \][/tex]
### Part c
We need to find the input [tex]\( x \)[/tex] when the output of the function is 24. That is, we need to solve for [tex]\( x \)[/tex] in the equation [tex]\( f(x) = 24 \)[/tex].
The function is:
[tex]\[ f(x) = x^2 - 1 \][/tex]
If [tex]\( f(x) = 24 \)[/tex]:
[tex]\[ x^2 - 1 = 24 \][/tex]
[tex]\[ x^2 = 24 + 1 \][/tex]
[tex]\[ x^2 = 25 \][/tex]
By taking the square root of both sides, we get:
[tex]\[ x = \pm 5 \][/tex]
So, there are two possible inputs for which the output is 24:
[tex]\[ x = 5 \][/tex] or [tex]\( x = -5 \)[/tex]
### Summary
- The output for [tex]\( f(4) \)[/tex] is [tex]\( 15 \)[/tex].
- The output for [tex]\( f(-1) \)[/tex] is [tex]\( 0 \)[/tex].
- The output for [tex]\( f(10) \)[/tex] is [tex]\( 99 \)[/tex].
- The inputs for which [tex]\( f(x) = 24 \)[/tex] are [tex]\( x = 5 \)[/tex] and [tex]\( x = -5 \)[/tex].
