4. at x = 3, the function given by f(x) = { x², x < 3; 6x - 9, x ≥ 3 is a undefined b continuous but not…

4. at x = 3, the function given by f(x) = { x², x < 3; 6x - 9, x ≥ 3 is a undefined b continuous but not differentiable c differentiable but not continuous d neither continuous nor differentiable e both continuous and differentiable

4. at x = 3, the function given by f(x) = { x², x < 3; 6x - 9, x ≥ 3 is a undefined b continuous but not differentiable c differentiable but not continuous d neither continuous nor differentiable e both continuous and differentiable

Answer

Explanation:

Step1: Check continuity

  • Left - hand limit: $\lim_{x\rightarrow3^{-}}f(x)=\lim_{x\rightarrow3^{-}}x^{2}=3^{2}=9$
  • Right - hand limit: $\lim_{x\rightarrow3^{+}}f(x)=\lim_{x\rightarrow3^{+}}(6x - 9)=6\times3-9=9$
  • Function value: $f(3)=6\times3 - 9=9$ Since $\lim_{x\rightarrow3^{-}}f(x)=\lim_{x\rightarrow3^{+}}f(x)=f(3) = 9$, the function is continuous at $x = 3$.

Step2: Check differentiability

  • Left - hand derivative: $f^{\prime}(x)=2x$ for $x\lt3$, so $f^{\prime}(3^{-})=\lim_{h\rightarrow0^{-}}\frac{f(3 + h)-f(3)}{h}=\lim_{h\rightarrow0^{-}}\frac{(3 + h)^{2}-9}{h}=\lim_{h\rightarrow0^{-}}\frac{9+6h+h^{2}-9}{h}=\lim_{h\rightarrow0^{-}}(6 + h)=6$
  • Right - hand derivative: $f^{\prime}(x)=6$ for $x\geq3$, so $f^{\prime}(3^{+})=\lim_{h\rightarrow0^{+}}\frac{f(3 + h)-f(3)}{h}=\lim_{h\rightarrow0^{+}}\frac{6(3 + h)-9-(6\times3 - 9)}{h}=\lim_{h\rightarrow0^{+}}\frac{18+6h-9 - 18 + 9}{h}=\lim_{h\rightarrow0^{+}}6=6$ Since $f^{\prime}(3^{-})=f^{\prime}(3^{+}) = 6$, the function is differentiable at $x = 3$.

Answer:

E. both continuous and differentiable