Q. 1 - Calculus

Question

1. True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: If f is both left and right continuous at x=c, then f is continuous at x=c.

(b) True or False: If f is continuous on the open interval (0,5), then f is continuous at every point in (0,5).

(c) True or False: If f is continuous on the closed interval [0,5], then f is continuous at every point in [0,5].

(d) True or False: If f is continuous on the interval (2,4), then f must have a maximum value and a minimum value on (2,4).

(e) True or False: If f(3)=-5 and f(9)=-2, then there must be a value c at which f(c)=-3.

(f) True or False: If f is continuous everywhere, and if f(-2)=3 and f(1)=2, then f(x) must have a root somewhere in (-2,1)

(g) True or False: If f is continuous everywhere, and if f(0)=-2 and f(4)=3, then f(x) must have a root somewhere in (0,4).

(h) True or False: If f(0)=f(6)=0 and f(2)>0, then f(x) is positive on the entire interval (0,6)

Step-by-Step Solution

Answer

(a) True

(b) True

(c) False

(d) False

(e) False

(f) False

(g) True

(h) False

Part (a) Step 1: Given information

If f is both left and right continuous at x=c, then f is continuous at x=c.

Part (a) Step 2: Simplification

From the definition of limit we have; a function f is left continuous at x=c if $\underset{x\to c^{-}}{\lim }f\left(x\right)=f\left(c\right)$ and right continuous at x=c if $\underset{x\to c^{-}}{\lim }f\left(x\right)=f\left(c\right)$.

Therefore, the statement is true.

Part (b) Step 1: Given information

If f is continuous on the open interval (0,5), then f is continuous at every point in (0,5).

Part (b) Step 2: Simplification

If a function f is continuous on the open interval (a, b), then f is continuous at every point in (a, b).

Therefore, the statement is true.

Part (c) Step 1: Given information

If f is continuous on the closed interval [0,5], then f is continuous at every point in [0,5].

Part (c) Step 2: Simplification

If a function f is continuous on the closed interval  [a, b],  then f is not continuous at every  point  in [a, b].

Therefore, the statement is false.

Part (d) Step 1: Given information

If f is continuous on the interval (2,4), then f must have a maximum value and a minimum value on (2,4).

Part (d) Step 2: Simplification

If f is continuous on the interval (a, b), then there is no any case that f have must a maximum value and a minimum value on (a, b).

Therefore, the statement is false.

Part (e) Step 1: Given information

If f(3)=-5 and f(9)=-2, then there must be a value c at which f(c)=-3.

Part (e) Step 2: Simplification

If f(3)=-5 and f(9)=-2 then without looking at the function we cannot say that there must be a value c at which f(c)=-3.

Therefore, the statement is false.

Part (f) Step 1: Given information

If f is continuous everywhere, and if f(-2)=3 and f(1)=2, then f(x) must have a root somewhere in (-2,1).

Part (f) Step 2: Simplification

If f is continuous everywhere, and if f(-2)=3 and f(1)=2 then without looking at the function we cannot say that f(x)must have a root somewhere in (0,4). If the graph does not intersect with x-axis in between 0 and 4 , then there is no root somewhere in (0,4).

Hence, the statement is false.

Part (g) Step 1: Given information

If f is continuous everywhere, and if f(0)=-2 and f(4)=3, then f(x) must have a root somewhere in (0,4).

Part (g) Step 2: Simplification

Here the two values are f(0)=-2 and f(4)=3. That means the graph must intersect the x axis from x=0 to x=4. Hence, there must be a root somewhere in (0,4).

Therefore, the statement is true.

Part (h) Step 1: Given information

If f(0)=f(6)=0 and f(2)>0, then f(x) is positive on the entire interval (0,6)

Part (h) Step 2: Simplification

Here the two value are f(0)=f(6)=0 and f(2)>0, then the graph must be curve up with intersection points at x=0,6. Hence, the function f(x) is negative on the interval of (0,6). Therefore, the statement is false.