QUESTION IMAGE
Question
$f(x) = \
$
let $f$ be the function given above. which of the following statements are true about $f$?
i. $\lim_{x \to 3} f(x)$ exists.
ii. $f$ is continuous at $x = 3$.
iii. $f$ is differentiable at $x = 3$.
options:
a none
b i only
c ii only
d i and ii only
e i, ii, and iii
Step1: Check Limit Existence (I)
To check if $\lim_{x\to 3} f(x)$ exists, we find the left - hand limit (LHL) and right - hand limit (RHL) as $x\to 3$.
For the left - hand limit, when $x\leq3$, $f(x)=x + 2$. So, $\lim_{x\to 3^{-}}f(x)=\lim_{x\to 3^{-}}(x + 2)$. Substitute $x = 3$ into $x + 2$, we get $3+2 = 5$.
For the right - hand limit, when $x>3$, $f(x)=4x - 7$. So, $\lim_{x\to 3^{+}}f(x)=\lim_{x\to 3^{+}}(4x - 7)$. Substitute $x = 3$ into $4x-7$, we get $4\times3 - 7=12 - 7 = 5$.
Since $\lim_{x\to 3^{-}}f(x)=\lim_{x\to 3^{+}}f(x) = 5$, the limit $\lim_{x\to 3}f(x)$ exists. So statement I is true.
Step2: Check Continuity at $x = 3$ (II)
A function $y = f(x)$ is continuous at $x=a$ if $\lim_{x\to a}f(x)=f(a)$.
We know that $\lim_{x\to 3}f(x) = 5$. Now, find $f(3)$. Since for $x\leq3$, $f(x)=x + 2$, then $f(3)=3 + 2=5$.
Since $\lim_{x\to 3}f(x)=f(3)=5$, the function $f(x)$ is continuous at $x = 3$. So statement II is true.
Step3: Check Differentiability at $x = 3$ (III)
A function $y = f(x)$ is differentiable at $x=a$ if the left - hand derivative (LHD) and right - hand derivative (RHD) at $x = a$ exist and are equal.
The derivative of a function $y = mx + c$ is $y^\prime=m$.
For the left - hand derivative (when $x\leq3$), $f(x)=x + 2$, so $f^\prime(x)=1$ for $x<3$. Then the left - hand derivative at $x = 3$, $f^\prime(3^{-}) = 1$.
For the right - hand derivative (when $x>3$), $f(x)=4x - 7$, so $f^\prime(x)=4$ for $x>3$. Then the right - hand derivative at $x = 3$, $f^\prime(3^{+}) = 4$.
Since $f^\prime(3^{-})=1
eq f^\prime(3^{+}) = 4$, the function $f(x)$ is not differentiable at $x = 3$. So statement III is false.
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D. I and II only