Let f be the function defined by f(x)= sqrt(x), 0 <or= x <or= 4. and f(x)= 6-x, 4 < x <or= 6
a. Is f continuous at x=4? justify.
b. Find the average rate of change of f(x) on the closed interval [0,6].
c. suppose the function g(x)= k sqrt(x), 0 <or= x <or= 4. and g(x)= mx+2, 4 < x <or= 6. where k and m are constants. find the values of k and m to make g differentiable at x=4? justify.
a. No, f is not continuous at x=4 because the left-hand limit and the right-hand limit of f(x) at x=4 are not equal. The left-hand limit of f(x) at x=4 is sqrt(4) = 2, while the right-hand limit of f(x) at x=4 is 6-4 = 2.
b. The average rate of change of f(x) on the closed interval [0,6] is (f(6)-f(0))/(6-0) = (6-sqrt(0))/6 = (6-0)/6 = 1.
c. To make g differentiable at x=4, the left-hand limit and the right-hand limit of g(x) at x=4 must be equal. This means that k sqrt(4) = 4m + 2. Solving for k and m, we get k = (4m+2)/sqrt(4) and m = (k sqrt(4)-2)/4.
a. To determine if f is continuous at x = 4, we need to check if the limit of f(x) as x approaches 4 from both the left and the right exists and is equal to f(4).
From the left side of 4, as x approaches 4, the function f(x) = sqrt(x) will approach sqrt(4) = 2.
From the right side of 4, as x approaches 4, the function f(x) = 6 - x will approach 6 - 4 = 2.
Since the limits from both sides equal 2, and f(4) also equals 2, we can conclude that f is continuous at x = 4.
b. To find the average rate of change of f(x) on the closed interval [0, 6], we can use the formula:
Average rate of change = (f(6) - f(0)) / (6 - 0)
In this case, we need to find f(6) and f(0).
For x ≤ 4, f(x) = sqrt(x). So, f(0) = sqrt(0) = 0.
For x > 4, f(x) = 6 - x. So, f(6) = 6 - 6 = 0.
Therefore, the average rate of change of f(x) on the closed interval [0, 6] is:
(0 - 0) / (6 - 0) = 0.
c. To make g differentiable at x = 4, we need to ensure that the function is continuous at x = 4 and that the derivative exists at x = 4.
First, let's check continuity. From the left side of 4, we have g(x) = k sqrt(x), which should approach g(4) as x approaches 4. From the right side of 4, we have g(x) = mx + 2, which should approach g(4) as x approaches 4.
From the left side of 4: k sqrt(4) = k(2) = 2k.
From the right side of 4: m(4) + 2 = 4m + 2.
To ensure continuity, we set 2k = 4m + 2 and solve for m in terms of k:
4m = 2k - 2
m = (2k - 2) / 4
m = (k - 1) / 2.
To ensure the derivative exists, we need the derivative of the two sides of the function to be equal at x = 4.
From the left side: g'(x) = (d/dx) (k sqrt(x)) = (k/2) (1/sqrt(x)).
From the right side: g'(x) = (d/dx) (mx + 2) = m.
Setting these two derivatives equal, we have:
(k/2) (1/sqrt(4)) = m,
(k/2) (1/2) = m,
(k/4) = m.
Now, we can equate the expressions for m and solve for k:
(k - 1) / 2 = k / 4,
2(k - 1) = k,
2k - 2 = k,
k = 2.
Substituting this value back into the expression for m, we have:
m = (k - 1) / 2 = (2 - 1) / 2 = 1 / 2.
Therefore, to make g differentiable at x = 4, the values of k and m are k = 2 and m = 1/2.
To answer the questions:
a. To determine if the function f is continuous at x = 4, we need to check if the left limit at x = 4 (as x approaches 4 from the left) is equal to the right limit at x = 4 (as x approaches 4 from the right), and if it is equal to the value of the function at x = 4.
Left limit at x = 4:
lim(x→4-) sqrt(x) = sqrt(4) = 2
Right limit at x = 4:
lim(x→4+) 6 - x = 6 - 4 = 2
Value of the function at x = 4:
f(4) = 6 - 4 = 2
Since the left limit, right limit, and the value of the function at x = 4 are all equal, f is continuous at x = 4.
b. To find the average rate of change of f(x) on the closed interval [0, 6], we need to evaluate the function at the endpoints and calculate the difference in the function values divided by the difference in the x-values.
Average rate of change of f(x) on [0, 6]:
= (f(6) - f(0)) / (6 - 0)
= (6 - 6 + 2 - 0) / 6
= 2 / 6
= 1/3
Therefore, the average rate of change of f(x) on [0, 6] is 1/3.
c. To make the function g differentiable at x = 4, we need to ensure that the left derivative at x = 4 (as x approaches 4 from the left) is equal to the right derivative at x = 4 (as x approaches 4 from the right).
Left derivative at x = 4:
lim(x→4-) k * sqrt(x) = k * sqrt(4) = 2k
Right derivative at x = 4:
lim(x→4+) m * x + 2 = m * 4 + 2 = 4m + 2
To make g differentiable at x = 4, the left derivative and the right derivative should be equal:
2k = 4m + 2
We do not have enough information to find the exact values of k and m, as there can be multiple valid values satisfying this equation. Additional conditions or information would be needed to determine the specific values of k and m.