Question

Pls help this is pretty urgent

Answer 1

Answer: See explanation

Step-by-step explanation:

10. a. x^2+4 doesn't equal 0.

x^2 doesn't equal -4. The square of a number is never negative

Hence, the denominator is never 0 so f has no restrictions, so f is continuous.

b. Yes.

4(x+4)=x^2+4

4x+16=x^2+4

0=x^2-4x-16+4

0=x^2-4x-12

0=(x-6)(x+2)

f(6) and f(-2) equal 0.25

c. When x approaches 0, f(x)-1 approaches 0 since f(x) approaches 1. The denominator approaches 0 too, so you get 0.0...01/0.0...01. Hence when x approaches 0, the lim when x approaches 0 is 1.

Answer 2

Answer:

(a)  See below

(b)  c = 6

(c)  1/4

Step-by-step explanation:

Given rational function:

$f(x)=\dfrac{x+4}{x^2+4}$

Part (a)

$\textsf{Rational function }f(x)=\dfrac{p(x)}{q(x)} \textsf{ is continuous for all real numbers}\\\\ \textsf{ except at points where }q(x)=0$

As x² + 4 ≥ 4, the function f(x) is continuous.

Part (b)

Set the function to 0.25 and solve for c:

$\begin{aligned} f(c) & = 0.25 \\\implies \dfrac{c+4}{c^2+4} & =0.25\\c+4 & = 0.25(c^2+4)\\ c + 4 & = 0.25c^2+1\\0.25c^2-c-3 & = 0\\ c^2-4c-12& = 0\\ c^2-6c+2c-12 & = 0\\c(c-6)+2(c-6) & = 0\\(c+2)(c-6) & = 0\\ \implies c & =-2, 6\end{aligned}$

Therefore, there exists a value of c in the interval [0, 100] such that f(c) = 0.25, where c = 6.

Part (c)

Find the new function:

$\begin{aligned}\dfrac{f(x)-1}{x} & = \dfrac{\dfrac{x+4}{x^2+4}-1}{x}\\& = \dfrac{\dfrac{x+4-x^2-4}{x^2+4}}{x}\\& = \dfrac{\dfrac{x-x^2}{x^2+4}}{x}\\& = \dfrac{x-x^2}{x(x^2+4)}\\& = \dfrac{1-x}{x^2+4}\end{aligned}$

Therefore:

$\displaystyle \lim_{x \to 0}\dfrac{f(x)-1}{x} =\lim_{x \to 0} \dfrac{1-x}{x^2+4}$

Plug in x = 0 to find the limit:

$\displaystyle \lim_{x \to 0} \dfrac{1-x}{x^2+4}=\dfrac{1-0}{0^2+4}=\dfrac{1}{4}$