Question

find each function value and the limit for f(x) = \frac{15 - 6x^{4}}{3 + x^{4}}. use -\infty or \infty where appropriate. (a) f(-10) (b) f(-20) (c) \lim_{x\to -\infty} f(x) (a) f(-10) = (round to the nearest thousandth as needed.)

Explanation:

Step1: Substitute x = - 10 into f(x)

$f(-10)=\frac{15 - 6\times(-10)^{4}}{3+(-10)^{4}}$
$=\frac{15-6\times10000}{3 + 10000}$
$=\frac{15-60000}{10003}$
$=\frac{-59985}{10003}\approx - 5.997$

Step2: Substitute x = - 20 into f(x)

$f(-20)=\frac{15-6\times(-20)^{4}}{3+(-20)^{4}}$
$=\frac{15-6\times160000}{3 + 160000}$
$=\frac{15 - 960000}{160003}$
$=\frac{-959985}{160003}\approx - 5.999$

Step3: Find the limit as x→ - ∞

Divide numerator and denominator by $x^{4}$:
$\lim_{x
ightarrow-\infty}f(x)=\lim_{x
ightarrow-\infty}\frac{\frac{15}{x^{4}}-6}{\frac{3}{x^{4}} + 1}$
As $x
ightarrow-\infty$, $\frac{15}{x^{4}}
ightarrow0$ and $\frac{3}{x^{4}}
ightarrow0$
So $\lim_{x
ightarrow-\infty}f(x)=\frac{0 - 6}{0 + 1}=-6$

Answer:

(A) $f(-10)\approx - 5.997$
(B) $f(-20)\approx - 5.999$
(C) $\lim_{x
ightarrow-\infty}f(x)=-6$