Question

F(x)=x2-10 and g(x)=11-x find (f-g)(x)(f-g)(10) and (f/g)(x) if they exist

Answer 1

Answer:

$(f - g)(x) = x^2 + x-21$

$(f - g)(10) = 89$

$(f/g)(x) = \frac{x^2 - 10}{11 - x}$

Step-by-step explanation:

Given

$f(x) = x^2 - 10$

$g(x) = 11 - x$

Solving (1): (f-g)(x)

$(f - g)(x) = f(x) - g(x)$

Substitute values for f(x) and g(x)

$(f - g)(x) = x^2 - 10 - (11 - x)$

Open Bracket

$(f - g)(x) = x^2 - 10 - 11 + x$

$(f - g)(x) = x^2 -21 + x$

Reorder

$(f - g)(x) = x^2 + x-21$

Solving (2): (f-g)(10)

In (1)

$(f - g)(x) = x^2 + x-21$

So:

$(f - g)(10) = 10^2 + 10-21$

$(f - g)(10) = 100 + 10-21$

$(f - g)(10) = 89$

Solving (3): (f/g)(x)

$(f/g)(x) = \frac{f(x)}{g(x)}$

Substitute values for f(x) and g(x)

$(f/g)(x) = \frac{x^2 - 10}{11 - x}$