Will give brainliest! use a table of values to find the solution to the equation ƒ(x ) = h(x ) where f(x)= \frac{1}{x+1} and h(x ) = 2x + 3. show your work

I'm assuming that you meant:

1

f(x) = and that you want to find the value of x at which f(x) = h(x).

x+1

Of course you could create a table for each f(x) and h(x), but setting f(x)=h(x) and solving for x algebraically would be faster and more efficient:

1

f(x) = = 2x + 3 = h(x). Then 1 = (x+1)(2x+3) = 2x^2 + 3x + 2x + 3

x+1

or 1 = 2x^2 + 5x + 3, or 2x^2 + 5x + 2 = 0.

This is a quadratic equation with a=2, b=5 and c=2. The discriminant is b^2-4ac, or 5^2-4(2)(2), OR 25-16= 9.

Thus, the roots are

-5 plus or minus sqrt(9)

x =

2(2)

-5 plus or minus 3

=

4

= {-1/2, -2}

Thus, f(x) = h(x) at both x=-1/2 and x= -2.

answer: graph 2

step-by-step explanation:

add 8+12 to get a simplified answer of 12r-20