Question

7 < x + 8
9 - x ≥ - 3
\frac{x}{2}< - 9
20 < 5x - 2
7x - 3x + 2 > x + 5x - 6
\frac{2x}{5}+6 ≤ 4
g(x) means “a number greater than x.” find a number to make each of these true.
g(4)=5
g(-2.3)= - 2
g(0)=1
fill in the missing numbers in the table for each function.
f(x)=x - 5
g(x)=x^{2}+1

Explanation:

Step1: Recall function - evaluation formula

For a function \(y = f(x)\), to find the value of the function at a given \(x\) - value, substitute the \(x\) - value into the function.

Step2: Evaluate \(f(x)\)

For \(f(x)=x - 5\), substitute each \(x\) value one - by - one. For example, when \(x = 100\), \(f(100)\) is obtained by replacing \(x\) with \(100\) in the expression \(x - 5\), so \(f(100)=100-5\).

Step3: Evaluate \(g(x)\)

For \(g(x)=x^{2}+1\), substitute each \(x\) value. For example, when \(x = 20\), \(g(20)\) is found by replacing \(x\) with \(20\) in the expression \(x^{2}+1\), so \(g(20)=20^{2}+1=400 + 1\).

Answer:

For \(f(x)=x - 5\):
When \(x = 100\), \(f(100)=100 - 5=95\)
When \(x = 10\), \(f(10)=10 - 5 = 5\)
When \(x = 4\), \(f(4)=4 - 5=-1\)
When \(x = 1\), \(f(1)=1 - 5=-4\)
When \(x = 0\), \(f(0)=0 - 5=-5\)
When \(x=-2\), \(f(-2)=-2 - 5=-7\)
When \(x=-5\), \(f(-5)=-5 - 5=-10\)

For \(g(x)=x^{2}+1\):
When \(x = 20\), \(g(20)=20^{2}+1=400 + 1=401\)
When \(x = 4\), \(g(4)=4^{2}+1=16 + 1=17\)
When \(x = 1\), \(g(1)=1^{2}+1=2\)
When \(x = 0\), \(g(0)=0^{2}+1=1\)
When \(x=-1\), \(g(-1)=(-1)^{2}+1=1 + 1=2\)
When \(x=-3\), \(g(-3)=(-3)^{2}+1=9 + 1=10\)
When \(x=-10\), \(g(-10)=(-10)^{2}+1=100 + 1=101\)