Question: Evaluate the function at the given values of the independent variable and simplify.
f left parenthesis x right parenthesis equals StartFraction 6 x squared minus 1 Over x squared EndFraction
(a) f left parenthesis 5 right parenthesis (b) f left parenthesis negative 5 right parenthesis (c) f left parenthesis negative x right parenthesis
1. **State the problem:**
We are given the function $$f(x) = \frac{6x^2 - 1}{x^2}$$ and need to evaluate it at three values: $$x=5$$, $$x=-5$$, and $$x=-x$$.
2. **Recall the formula:**
$$f(x) = \frac{6x^2 - 1}{x^2}$$
3. **Evaluate (a) $$f(5)$$:**
$$f(5) = \frac{6(5)^2 - 1}{(5)^2} = \frac{6 \times 25 - 1}{25} = \frac{150 - 1}{25} = \frac{149}{25}$$
4. **Evaluate (b) $$f(-5)$$:**
Since $$x^2$$ is the same for $$x=5$$ and $$x=-5$$, we have:
$$f(-5) = \frac{6(-5)^2 - 1}{(-5)^2} = \frac{6 \times 25 - 1}{25} = \frac{149}{25}$$
5. **Evaluate (c) $$f(-x)$$:**
Substitute $$-x$$ into the function:
$$f(-x) = \frac{6(-x)^2 - 1}{(-x)^2} = \frac{6x^2 - 1}{x^2}$$
Note that $$(-x)^2 = x^2$$, so the function remains the same.
**Final answers:**
(a) $$\frac{149}{25}$$
(b) $$\frac{149}{25}$$
(c) $$\frac{6x^2 - 1}{x^2}$$