Consider the function f : R → R, defined as f(x) = ⎧⎪⎪⎪⎪�

1.	Consider the function f : R → R, defined as f(x) = ⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩x2−x+3,xϵ(−∞,3)∩Qx+a,xϵ(−∞,2)−Q2x+1,xϵ(2,3)−Q9 tan(π x12),xϵ[3,6] f(2+)=22+1=5 through irrational f(2−)=2+a through rational f(2) = 4 - 2 + 3 = 5 Hence for continuity at x = 2 2 + a = 5 ⇒ a = 3. At x=3For x=3+; f(x)=9tanπx12⇒f(3+)=9tan3π12=9f′(x)=9π12sec2πx12⇒f′(3+)=9π12sec23π12=3π2≈4.71For x=3−; f(x)={x2−x+3; xϵQ2x+1; xϵR−Q⇒f(3−)={9; xϵQ9; xϵR−Qf′(x)={2x−1; xϵQ2xln2; xϵR−Q⇒f′(3−)={5; x∈Q8ln2; x∈R−Q≈{5; xϵQ5.54; xϵR−Q Therefore, continuous but not differentiable at x=3.
Answer» Consider the function f : R $\to$ R, defined as f(x) = $⎧ ⎪⎪⎪⎪⎪⎪ ⎨ ⎪⎪⎪⎪⎪⎪ ⎩ \begin{matrix} x^{2} - x + 3, x ϵ (- \infty, 3) \cap Q x + a, x ϵ (- \infty, 2) - Q 2^{x} + 1, x ϵ (2, 3) - Q 9 t a n (\frac{π x}{12}), x ϵ [3, 6] \end{matrix}$ $f (2^{+}) = 2^{2} + 1 = 5$ through irrational $f (2^{-}) = 2 + a$ through rational f(2) = 4 - 2 + 3 = 5 Hence for continuity at x = 2 2 + a = 5 $\Rightarrow$ a = 3. $At x = 3 For x = 3^{+}; f (x) = 9 tan \frac{π x}{12} \Rightarrow f (3^{+}) = 9 tan \frac{3 π}{12} = 9 f^{'} (x) = 9 \frac{π}{12} {sec}^{2} \frac{π x}{12} \Rightarrow f^{'} (3^{+}) = 9 \frac{π}{12} {sec}^{2} \frac{3 π}{12} = \frac{3 π}{2} \approx 4.71 For x = 3^{-}; f (x) = {\begin{matrix} x^{2} - x + 3; & x ϵ Q 2^{x} + 1; & x ϵ R - Q \end{matrix} \Rightarrow f (3^{-}) = {\begin{matrix} 9; & x ϵ Q 9; & x ϵ R - Q \end{matrix} f^{'} (x) = {\begin{matrix} 2 x - 1; & x ϵ Q 2^{x} ln 2; & x ϵ R - Q \end{matrix} \Rightarrow f^{'} (3^{-}) = {\begin{matrix} 5; & x \in Q 8 ln 2; & x \in R - Q \end{matrix} \approx {\begin{matrix} 5; & x ϵ Q 5.54; & x ϵ R - Q \end{matrix}$ Therefore, continuous but not differentiable at x=3.

Consider the function f : R → R, defined as f(x) = ⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩x2−x+3,xϵ(−∞,3)∩Qx+a,xϵ(−∞,2)−Q2x+1,xϵ(2,3)−Q9 tan(π x12),xϵ[3,6] f(2+)=22+1=5 through irrational f(2−)=2+a through rational f(2) = 4 - 2 + 3 = 5 Hence for continuity at x = 2 2 + a = 5 ⇒ a = 3. At x=3For x=3+; f(x)=9tanπx12⇒f(3+)=9tan3π12=9f′(x)=9π12sec2πx12⇒f′(3+)=9π12sec23π12=3π2≈4.71For x=3−; f(x)={x2−x+3; xϵQ2x+1; xϵR−Q⇒f(3−)={9; xϵQ9; xϵR−Qf′(x)={2x−1; xϵQ2xln2; xϵR−Q⇒f′(3−)={5; x∈Q8ln2; x∈R−Q≈{5; xϵQ5.54; xϵR−Q Therefore, continuous but not differentiable at x=3.

Answer»

Consider the function f : R $\to$ R, defined as f(x) = $⎧ ⎪⎪⎪⎪⎪⎪ ⎨ ⎪⎪⎪⎪⎪⎪ ⎩ \begin{matrix} x^{2} - x + 3, x ϵ (- \infty, 3) \cap Q x + a, x ϵ (- \infty, 2) - Q 2^{x} + 1, x ϵ (2, 3) - Q 9 t a n (\frac{π x}{12}), x ϵ [3, 6] \end{matrix}$

$f (2^{+}) = 2^{2} + 1 = 5$

through irrational

$f (2^{-}) = 2 + a$

through rational

f(2) = 4 - 2 + 3 = 5

Hence for continuity at x = 2

2 + a = 5 $\Rightarrow$ a = 3.

$At x = 3 For x = 3^{+}; f (x) = 9 tan \frac{π x}{12} \Rightarrow f (3^{+}) = 9 tan \frac{3 π}{12} = 9 f^{'} (x) = 9 \frac{π}{12} {sec}^{2} \frac{π x}{12} \Rightarrow f^{'} (3^{+}) = 9 \frac{π}{12} {sec}^{2} \frac{3 π}{12} = \frac{3 π}{2} \approx 4.71 For x = 3^{-}; f (x) = {\begin{matrix} x^{2} - x + 3; & x ϵ Q 2^{x} + 1; & x ϵ R - Q \end{matrix} \Rightarrow f (3^{-}) = {\begin{matrix} 9; & x ϵ Q 9; & x ϵ R - Q \end{matrix} f^{'} (x) = {\begin{matrix} 2 x - 1; & x ϵ Q 2^{x} ln 2; & x ϵ R - Q \end{matrix} \Rightarrow f^{'} (3^{-}) = {\begin{matrix} 5; & x \in Q 8 ln 2; & x \in R - Q \end{matrix} \approx {\begin{matrix} 5; & x ϵ Q 5.54; & x ϵ R - Q \end{matrix}$
Therefore, continuous but not differentiable at x=3.

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