Let f(x) = [x] + |1 – x|, – 1 ≤ x ≤ 3 and [x] is the largest integer not exceeding x. The number of points in [–1, 3] where f is not continuous is

Let f(x)=∣∣2x2+5∣∣x|−3|,x∈R. If m and n denote the number of points where f is not continuous and not differentiable respectively, then m+n is equal to :

Which function has a discontinuity at x=3?Responsesf(x)={3x+1 for x<3x2+1 for x≥3𝑓(𝑥)={3𝑥+1 𝑓𝑜𝑟 𝑥<3𝑥2+1 𝑓𝑜𝑟 𝑥≥3f(x)={3x+1 for x<3x2+1 for x≥3𝑓(𝑥)={3𝑥+1 𝑓𝑜𝑟 𝑥<3𝑥2+1 𝑓𝑜𝑟 𝑥≥3f(x)=|x−3|+2𝑓(𝑥)=|𝑥−3|+2f of x is equal to start absolute value x minus 3 end absolute value plus 2f(x)=x−3x2𝑓(𝑥)=𝑥−3𝑥2f of x is equal to the fraction with numerator x minus 3 and denominator x squaredf(x)=x+2x2−9

How many discontinuities are in the function f(x), where f(x) = -1 when x < -1, and f(x) = x for values of x ≥ -1?

(1) f (x) = x2 − 4x + 3; [1, 3](2) f (x) = x3 − x; [−1, 1](3) f (x) = x3 − 9x; [−3, 3]Solution(1) f (x) = x2 − 4x + 3; [1, 3].f (1) = 1 − 4 + 3 = 0f (3) = 9 − 12 + 3 = 0The polynomial f (x) is continuous and dierentiable on [1, 3].Therefore the hypotheses of Rolle's Theorem are satised.

Consider the piecewise functionf (x) =x + 1, if x < −21, if − 2 ≤ x ≤ 1x2, if x > 1.(i) Find limx→−2 f (x) if it exists.(ii) Show that f is continuous at x = 1.(iii) Sketch the graph of f (x)