Find the relative extrema, if any, of the function. Use the second derivative test, if applicable. (If an answer does not exist, enter DNE.)
f(t) = 7 t + 3/t
relative maximum (x, y) =
relative minimum (x, y) =
f' = 7-3/t^2
f'=0 when t=±√(3/7)
f" = 6/t^3
f" > 0 when t>0 so min at t=√(3/7)
f" < 0 when t<0 so max at t=-√(3/7)
To find the relative extrema of the function f(t) = 7t + 3/t, we first need to take the derivative and set it equal to zero to find any critical points.
1. Find the derivative of f(t):
f'(t) = 7 - 3/t^2
2. Set f'(t) = 0 and solve for t:
7 - 3/t^2 = 0
3/t^2 = 7
t^2/3 = 1/7
t^2 = 3/7
t = ±√(3/7)
We have found the critical points at t = ±√(3/7).
Next, we need to determine whether these critical points are relative extrema using the second derivative test. If the second derivative is positive at a critical point, it is a relative minimum. If the second derivative is negative at a critical point, it is a relative maximum.
3. Find the second derivative of f(t):
f''(t) = 6/t^3
4. Evaluate the second derivative at the critical points:
f''(√(3/7)) = 6 / (√(3/7))^3 = 6 / (∛(3/7))^2 = 6 / (∛(3^2 / 7^2)) = 6 / (∛9 / 49) = 6 * 49 / ∛9 = 294 / ∛9
f''(-√(3/7)) = 6 / (-√(3/7))^3 = 6 / (-∛(3/7))^2 = 6 / (-∛(3^2 / 7^2)) = 6 / (-∛9 / 49) = -6 * 49 / ∛9 = -294 / ∛9
Since both f''(√(3/7)) and f''(-√(3/7)) are nonzero, the second derivative test is inconclusive.
Therefore, there are no relative extrema for the function f(t) = 7t + 3/t.
To find the relative extrema of the function f(t) = 7t + 3/t, we need to take the derivative of the function, set it equal to zero, and then determine the nature of the critical points using the second derivative test.
Step 1: Find the derivative of f(t).
To find the derivative f'(t), we need to apply the power rule for differentiation and the quotient rule.
f'(t) = 7 - 3/t^2
Step 2: Set the derivative equal to zero and solve for t.
To find the critical points, we set f'(t) equal to zero and solve for t.
7 - 3/t^2 = 0
Step 3: Solve for t.
To solve the equation, we can rearrange it as follows:
7 = 3/t^2
Multiply both sides of the equation by t^2 to get:
7t^2 = 3
Divide both sides of the equation by 7 to isolate t^2:
t^2 = 3/7
Take the square root of both sides to solve for t:
t = ±√(3/7)
So the critical points are t = √(3/7) and t = -√(3/7).
Step 4: Find the second derivative of f(t).
To apply the second derivative test, we need to find the second derivative of f(t), denoted as f''(t).
f''(t) = 6/t^3
Step 5: Analyze the nature of the critical points using the second derivative test.
We evaluate the second derivative at each critical point to classify them as relative maximum or minimum.
At t = √(3/7):
f''(√(3/7)) = 6/(√(3/7))^3 = 6/(3/7√(3)) = 6(7√(3))/(3√(3)) = 14√(3)
Since f''(√(3/7)) > 0, the function has a relative minimum at t = √(3/7).
At t = -√(3/7):
f''(-√(3/7)) = 6/(-√(3/7))^3 = 6/(-3/7√(3)) = -6(7√(3))/(3√(3)) = -14√(3)
Since f''(-√(3/7)) < 0, the function has a relative maximum at t = -√(3/7).
Therefore, the relative extrema are as follows:
Relative maximum: (-√(3/7), f(-√(3/7)))
Relative minimum: (√(3/7), f(√(3/7)))
To find the corresponding y-values, substitute the critical points into the original function:
f(-√(3/7)) = 7(-√(3/7)) + 3/(-√(3/7))
f(-√(3/7)) = -7√(3) - 3√(7)
f(√(3/7)) = 7(√(3/7)) + 3/(√(3/7))
f(√(3/7)) = 7√(3) + 3√(7)
Therefore, the relative maximum is (-√(3/7), -7√(3) - 3√(7)), and the relative minimum is (√(3/7), 7√(3) + 3√(7)).