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21 Find the values of x for which (a) f(x)={x2+10 if x=0 if x=0 is continuous. (b) f(x)=x2−2xx2−3x+2 (c) g(x)=∣x+1∣ (d) f(x)={x2x−1 if x≤1 if x>1

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Answer:

Let's analyze each function separately to find the values of x for which they are continuous:

(a) f(x) = {x^2 + 10 if x ≠ 0, 0 if x = 0}

To check for continuity at x = 0, we need to evaluate the limit as x approaches 0 from both the left and the right and see if they match the function value at x = 0.

Limit as x approaches 0 from the left:

lim (x → 0-) (x^2 + 10) = 10

Limit as x approaches 0 from the right:

lim (x → 0+) (x^2 + 10) = 10

Function value at x = 0:

f(0) = 0

Since the limit from both sides matches the function value at x = 0, the function is continuous at x = 0.

(b) f(x) = x^2 - (2x / (x^2 - 3x + 2))

To check for continuity, we need to find the values of x that make the denominator zero (potential discontinuities) and then check the limit as x approaches those values.

The denominator is x^2 - 3x + 2, which factors as (x - 1)(x - 2).

Setting the denominator equal to zero and solving for x:

x - 1 = 0 => x = 1

x - 2 = 0 => x = 2

Potential discontinuities occur at x = 1 and x = 2. Let's check the limits at these points:

Limit as x approaches 1:

lim (x → 1) [x^2 - (2x / (x^2 - 3x + 2))] = 1^2 - (2 * 1 / (1^2 - 3 * 1 + 2)) = 1 - (2 / (1 - 3 + 2)) = 1 - (2 / 0) (undefined)

Limit as x approaches 2:

lim (x → 2) [x^2 - (2x / (x^2 - 3x + 2))] = 2^2 - (2 * 2 / (2^2 - 3 * 2 + 2)) = 4 - (4 / (4 - 6 + 2)) = 4 - (4 / 0) (undefined)

The function is not continuous at x = 1 and x = 2 due to the division by zero in the denominator.

(c) g(x) = |x + 1|

The absolute value function is continuous everywhere, so g(x) is continuous for all real numbers.

(d) f(x) = {x^2 if x ≤ 1, x - 1 if x > 1}

To check for continuity, we need to evaluate the limit as x approaches 1 from both the left and the right and see if they match the function value at x = 1.

Limit as x approaches 1 from the left:

lim (x → 1-) (x^2) = 1^2 = 1

Limit as x approaches 1 from the right:

lim (x → 1+) (x - 1) = 1 - 1 = 0

Function value at x = 1:

f(1) = 1^2 = 1

Since the limit from the left is not equal to the limit from the right, the function is not continuous at x = 1.

In summary:

(a) f(x) is continuous for all real numbers.

(b) f(x) is not continuous at x = 1 and x = 2.

(c) g(x) is continuous for all real numbers.

(d) f(x) is not continuous at x = 1.

Explanation:

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