find each function value and the limit for f(x) = \\frac{15 - 6x^{4}}{3 + x^{4}}. use -\\infty or \\infty…

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.)

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.)

Answer

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 $x^{4}$: $\lim_{x\rightarrow-\infty}f(x)=\lim_{x\rightarrow-\infty}\frac{\frac{15}{x^{4}}-6}{\frac{3}{x^{4}} + 1}$ As $x\rightarrow-\infty$, $\frac{15}{x^{4}}\rightarrow0$ and $\frac{3}{x^{4}}\rightarrow0$ So $\lim_{x\rightarrow-\infty}f(x)=\frac{0 - 6}{0 + 1}=-6$

Answer:

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