Question

$f\left(x\right)=[\begin{array}{c}\frac{1}{4}(3x^2+1)-\infty ⟨x\le 1\\ 5-4x1⟨x⟨4\\ 4-x4\le x⟨\infty \end{array}$

• A. continuous at x =1 & x=4
• B. continuous at x = 1 , discontinuous at x = 4
• C. continuous at x = 4 , discontinuous at x = 1
• D. discontinuous at x = 4 & x = 1

Answer 1

The correct option is B continuous at x = 1 , discontinuous at x = 4

$f\left(1^{-}\right)=\frac{1}{4}(3.1^2+1)=1f\left(1^{+}\right)=5-4.1=1f\left(1^{-}\right)=f\left(1^{+}\right)$

so $f\left(x\right)$ continuous of $x=1$

$f\left(4^{-}\right)=5-4.4=5-16=-11f\left(4^{+}\right)=4-4=0f\left(4^{-}\right)\ne f\left(4^{+}\right)$

so $f\left(x\right)$ discontinuous at $x=4$