Analyze the function. f(x)={12x+12ifx&lt;22x+11ifx≥2 A.) Is f(x) continuous at x=2? B.) Is f(x) differentiable at x=2?

Step 1 f(x)={12x+12ifx&lt;22x+11ifx≥2 We know thatm if f(x) is continuous at x=2. Then Rlimx→2f(x)=Llimx→2f(x)=limx→2f(x) i.e. limx→2+f(x)=limx→2−f(x)=limx→2f(x) Now limx→2+=limx→2+(2x+11) =limx→2+2x+limx→2+11 =2(2)+11 =2+11 =13 limx→2−f(x)=limx→2−(x2+12) =22+11=1+11=12 Hence Rlim≠Llim ⇒f(x) is not continuous at x=2 We know that, if a function is differetiable at a point, then it must be continous at that point. Since f(x) is not Continuou at x=2 Thus f(x) is not differentiable at x=2 Answer: Part 1: no Part 2: no

Solved: "Analyze the function. f(x)={12x+12ifx<22x+11ifx≥2 A.) Is f(x) continuous..."

Randy Mejia

Answered question

2022-02-13

Analyze the function.

f (x) = {\begin{array}{ccc} \frac{1}{2} x + 12 & if & x < 2 \\ \sqrt{2 x} + 11 & if & x \geq 2 \end{array}

A.) Is f(x) continuous at

x = 2

?
B.) Is f(x) differentiable at

x = 2

Answer & Explanation

mikedejoyacqs

Beginner2022-02-14Added 11 answers

Step 1
$f (x) = {\begin{array}{ccc} \frac{1}{2} x + 12 & if & x < 2 \\ \sqrt{2 x} + 11 & if & x \geq 2 \end{array}$
We know thatm if f(x) is continuous at $x = 2$ . Then
$R lim_{x \to 2} f (x) = L lim_{x \to 2} f (x) = lim_{x \to 2} f (x)$
i.e. $lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{-}} f (x) = lim_{x \to 2} f (x)$
Now
$lim_{x \to 2^{+}} = lim_{x \to 2^{+}} (\sqrt{2 x} + 11)$
$= lim_{x \to 2^{+}} \sqrt{2 x} + lim_{x \to 2^{+}} 11$
$= \sqrt{2 (2)} + 11$
$= 2 + 11$
$= 13$
$lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} (\frac{x}{2} + 12)$
$= \frac{2}{2} + 11 = 1 + 11 = 12$
Hence $R lim \neq L lim$
$\Rightarrow f (x)$ is not continuous at $x = 2$
We know that, if a function is differetiable at a point, then it must be continous at that point.
Since f(x) is not Continuou at $x = 2$
Thus f(x) is not differentiable at $x = 2$
Answer: Part 1: no
Part 2: no

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