f:x⟼1x2−9,x≠±3g:x⟼x+2h:x⟼1x+23 ,   x≠−23Solve  fg(x)=h(x)

To solve the equation fg(x) = h(x), we first need to find the expressions for fg(x) and h(x).

The function fg(x) is the composition of f and g, which means we substitute g(x) into f(x).

g(x) = x + 2

Substitute g(x) into f(x):

f(g(x)) = 1/(g(x))^2 - 9 = 1/(x + 2)^2 - 9

The function h(x) is given by:

h(x) = 1/(x + 23)

Now, we set fg(x) equal to h(x) and solve for x:

1/(x + 2)^2 - 9 = 1/(x + 23)

To solve this equation, we first get rid of the fractions by multiplying through by (x + 2)^2 * (x + 23):

(x + 23) - 9(x + 2)^2 = (x + 2)^2

Expand and simplify:

x + 23 - 9x^2 - 36x - 72 = x^2 + 4x + 4

Combine like terms:

10x^2 + 40x + 53 = 0

This is a quadratic equation, which we can solve using the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / (2a)

Here, a = 10, b = 40, and c = 53. Substituting these values in gives:

x = [-40 ± sqrt((40)^2 - 41053)] / (2*10) x = [-40 ± sqrt(1600 - 2120)] / 20 x = [-40 ± sqrt(-520)] / 20

Since the square root of a negative number is not a real number, there are no real solutions to this equation. Therefore, fg(x) = h(x) has no solutions.