The number of discontinuities in mathbb{R} fo | vedclass

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(A) Given the function $f(x) = \frac{x-1}{x^3+6x^2+11x+6}$.First,we factorize the denominator $x^3+6x^2+11x+6$.By testing roots,we find that $x = -1$

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$3$

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(A) Given the function $f(x) = \frac{x-1}{x^3+6x^2+11x+6}$.First,we factorize the denominator $x^3+6x^2+11x+6$.By testing roots,we find that $x = -1$ is a root since $(-1)^3 + 6(-1)^2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0$.Dividing the polynomial by $(x+1)$,we get $x^2+5x+6$,which further factorizes into $(x+2)(x+3)$.Thus,$f(x) = \frac{x-1}{(x+1)(x+2)(x+3)}$.$A$ rational function is discontinuous where its denominator is zero.Setting the denominator to zero: $(x+1)(x+2)(x+3) = 0$.This gives the points of discontinuity as $x = -1, -2, -3$.Since there are $3$ such points,the number of discontinuities in $\mathbb{R}$ is $3$.

The number of discontinuities in $\mathbb{R}$ for the function $f(x)=\frac{x-1}{x^3+6x^2+11x+6}$ is

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