If f(x) \(\begin{cases}-x^2; &-1\leq x < 0\\ 4x - 3;& 0 < x \leq1\\5x

1.	If f(x) \(\begin{cases}-x^2; &-1\leq x < 0\\ 4x - 3;& 0 < x \leq1\\5x^2 - 4x; & 1 < x \leq 2\end{cases}\) then test the continuity of f (x) is closed interval [- 1, 2].
Answer» Here, we will test the continuity of function at x = 0 and 0 ∈ [-1,2]. Left hand limit f(0 – 0) = lim_h→0 f(0 – h) = lim_h→0 (0 – h)² = lim_h→0 h² = 0 Right hand limit f (0 + 0) = lim_h→0 f(0 + h) = lim_h→0 4(0 + h) – 3 = lim_h→0 Ah – 3 = 0 – 3 = – 3 ∴ f (0 – 0) ≠ f(0 + 0) LHL ≠ RHL So, function is not continuous at x = 0 and x ∈ [- 1, 2] At x = 1 Left hand limit f(1 – h) = limh→0 4(1 – h) – 3 = lim_h→0 4 – 3 – Ah =4 – 3 – 0 = 1 Right hand limit f(1 + h) = lim_h→0 5(1 + h)² – 4 (1 + h) = lim_h→0 5(1 + h² + 2h) – (4 + Ah) = lim_h→0 5h² +10h + 5 – 4 – 4h = 5 × 0 + 10 × 0 + 1 – 4(0) = 1 Value of function at x = 1 f(1) = 4 × 1 – 3 = 1 ∵ lim_h→0 f(1 – A) = f(1) = lim_h→0 f(1 + h) ∴Function is continuous at x = 1. Hence, function is not continuous in interval [-1, 2],

If f(x) \(\begin{cases}-x^2; &-1\leq x < 0\\ 4x - 3;& 0 < x \leq1\\5x^2 - 4x; & 1 < x \leq 2\end{cases}\) then test the continuity of f (x) is closed interval [- 1, 2].

Answer»

Here, we will test the continuity of function at x = 0 and 0 ∈ [-1,2].

Left hand limit

f(0 – 0) = lim_h→0 f(0 – h)

= lim_h→0 (0 – h)²

= lim_h→0 h²

= 0

Right hand limit

f (0 + 0) = lim_h→0 f(0 + h)

= lim_h→0 4(0 + h) – 3

= lim_h→0 Ah – 3

= 0 – 3

= – 3

∴ f (0 – 0) ≠ f(0 + 0)

LHL ≠ RHL

So, function is not continuous at x = 0 and x ∈ [- 1, 2]

At x = 1

Left hand limit

f(1 – h) = limh→0 4(1 – h) – 3

= lim_h→0 4 – 3 – Ah

=4 – 3 – 0 = 1

Right hand limit

f(1 + h) = lim_h→0 5(1 + h)² – 4 (1 + h)

= lim_h→0 5(1 + h² + 2h) – (4 + Ah)

= lim_h→0 5h² +10h + 5 – 4 – 4h

= 5 × 0 + 10 × 0 + 1 – 4(0)

= 1

Value of function at x = 1

f(1) = 4 × 1 – 3 = 1

∵ lim_h→0 f(1 – A) = f(1) = lim_h→0 f(1 + h)

∴Function is continuous at x = 1.

Hence, function is not continuous in interval [-1, 2],