let f(x) = {7x, x ≤ 5; x², x > 5. find each point of discontinuity of f, and for each give the value of the…

let f(x) = {7x, x ≤ 5; x², x > 5. find each point of discontinuity of f, and for each give the value of the point of discontinuity and evaluate the indicated one - sided limits. note: you use i for ∞ and -i for -∞. if you have more than one point, give them in numerical order, from smallest to largest. if you have extra boxes, fill each in with an x. point 1: c = lim x→c⁻ f(x) = lim x→c⁺ f(x) = point 2: c = lim x→c⁻ f(x) = lim x→c⁺ f(x) = point 3: c = lim x→c⁻ f(x) = lim x→c⁺ f(x) =

let f(x) = {7x, x ≤ 5; x², x > 5. find each point of discontinuity of f, and for each give the value of the point of discontinuity and evaluate the indicated one - sided limits. note: you use i for ∞ and -i for -∞. if you have more than one point, give them in numerical order, from smallest to largest. if you have extra boxes, fill each in with an x. point 1: c = lim x→c⁻ f(x) = lim x→c⁺ f(x) = point 2: c = lim x→c⁻ f(x) = lim x→c⁺ f(x) = point 3: c = lim x→c⁻ f(x) = lim x→c⁺ f(x) =

Answer

Explanation:

Step1: Recall the definition of continuity

A function $y = f(x)$ is continuous at $x = c$ if $\lim_{x\rightarrow c^{-}}f(x)=\lim_{x\rightarrow c^{+}}f(x)=f(c)$. We need to check the one - sided limits at the point where the function changes its rule, which is $x = 5$.

Step2: Calculate the left - hand limit as $x\rightarrow5^{-}$

For $x\rightarrow5^{-}$ (values of $x$ less than 5), $f(x)=7x$. So, $\lim_{x\rightarrow5^{-}}f(x)=\lim_{x\rightarrow5^{-}}7x$. Using the direct - substitution property of limits, we substitute $x = 5$ into $7x$. So, $\lim_{x\rightarrow5^{-}}7x=7\times5 = 35$.

Step3: Calculate the right - hand limit as $x\rightarrow5^{+}$

For $x\rightarrow5^{+}$ (values of $x$ greater than 5), $f(x)=x^{2}$. So, $\lim_{x\rightarrow5^{+}}f(x)=\lim_{x\rightarrow5^{+}}x^{2}$. Using the direct - substitution property of limits, we substitute $x = 5$ into $x^{2}$. So, $\lim_{x\rightarrow5^{+}}x^{2}=5^{2}=25$. Since $\lim_{x\rightarrow5^{-}}f(x)=35$ and $\lim_{x\rightarrow5^{+}}f(x)=25$, the function $f(x)$ is discontinuous at $x = 5$.

Answer:

Point 1: $C = 5$ $\lim_{x\rightarrow5^{-}}f(x)=35$ $\lim_{x\rightarrow5^{+}}f(x)=25$ Point 2: $C=x$ $\lim_{x\rightarrow x^{-}}f(x)=x$ $\lim_{x\rightarrow x^{+}}f(x)=x$ Point 3: $C=x$ $\lim_{x\rightarrow x^{-}}f(x)=x$ $\lim_{x\rightarrow x^{+}}f(x)=x$