or what value of the constant c is the function f continuous on (−∞,∞) where

f(x)={cx+7cx2−7 if x∈(−∞,3] if x∈(3,∞)

To find the value of the constant c that makes the function f continuous on (-∞, ∞), we need to ensure that the left-hand limit and the right-hand limit of f at x = 3 are equal.

First, let's find the left-hand limit of f as x approaches 3:
lim(x->3-) f(x) = lim(x->3-) (cx + 7cx^2 - 7)

Since f(x) is defined as cx + 7cx^2 - 7 for x ≤ 3, we can simply substitute 3 into the function:
lim(x->3-) f(x) = c(3) + 7c(3)^2 - 7
= 3c + 63c - 7
= 66c - 7

Next, let's find the right-hand limit of f as x approaches 3:
lim(x->3+) f(x) = lim(x->3+) (cx + 7cx^2 - 7)

Since f(x) is defined as 0 for x > 3, we can substitute 3 into the function:
lim(x->3+) f(x) = 0

To ensure that the function is continuous at x = 3, we need the left-hand limit and the right-hand limit to be equal:
lim(x->3-) f(x) = lim(x->3+) f(x)

Therefore, we have: 66c - 7 = 0

To solve for c, we can add 7 to both sides and then divide by 66:
66c = 7
c = 7/66

So, the value of the constant c that makes the function f continuous on (-∞, ∞) is c = 7/66.

To determine the value of the constant c that makes the function f continuous on the interval (-∞, ∞), we need to ensure that the left and right limits of the function at the point of transition, which is x = 3 in this case, are equal.

First, let's find the left limit as x approaches 3. We evaluate f(x) as x approaches 3 from the left side:

lim (x → 3-) f(x) = lim (x → 3-) (cx + 7cx^2 - 7)

Since the function is defined differently for x ≤ 3 and x > 3, we only consider the expression when x ≤ 3:

lim (x → 3-) f(x) = lim (x → 3-) (cx + 7cx^2 - 7) = 3c + 7(3c)^2 - 7

Next, let's find the right limit as x approaches 3. We evaluate f(x) as x approaches 3 from the right side:

lim (x → 3+) f(x) = lim (x → 3+) (cx + 7cx^2 - 7)

Since the function is defined differently for x ≤ 3 and x > 3, we only consider the expression when x > 3:

lim (x → 3+) f(x) = lim (x → 3+) (cx + 7cx^2 - 7) = c(3) + 7c(3)^2 - 7

To make f(x) continuous at x = 3, the limit from the left side should be equal to the limit from the right side. Therefore, we set them equal to each other:

3c + 7(3c)^2 - 7 = c(3) + 7c(3)^2 - 7

Simplifying the equation, we have:

3c + 63c^2 - 7 = 3c + 63c^2 - 7

The terms with c cancel out, and we are left with 0 = 0. This indicates that the equation is true for any value of c. Therefore, there is no specific value of c that makes the function f continuous on the interval (-∞, ∞). The function is continuous regardless of the value of c.