MATH SOLVE

10 months ago

Q:
# Prove that x²+6x+18 is always greater than[tex]2 - \frac{1}{x} [/tex]For any value of x.

Accepted Solution

A:

One way to solve this would be to graph both y=x^2+6x+18 and y=2 - 1/x. If the graph of the polynomial is always higher up (above) the graph of y=2-1/x, that alone is sufficient cause to state that the poly is always greater than 2-1/x.

You could also do this algebraically: write the inequality

x^2+6x+18 > 2 -1/x. This can be rewritten as x^2+6x+18-2+1/x > 0, or

x^2+6x+16+1/x > 0

Try x=-3. Then 9-18+16-1/3 = 6 2/3, which is greater than 0.

Or you could graph x^2+6x+16+1/x by hand or on a calculator. Is the graph always above the x-axis? If so, x²+6x+18 is always greater than 2-1/x.

You could also do this algebraically: write the inequality

x^2+6x+18 > 2 -1/x. This can be rewritten as x^2+6x+18-2+1/x > 0, or

x^2+6x+16+1/x > 0

Try x=-3. Then 9-18+16-1/3 = 6 2/3, which is greater than 0.

Or you could graph x^2+6x+16+1/x by hand or on a calculator. Is the graph always above the x-axis? If so, x²+6x+18 is always greater than 2-1/x.