Application Of Derivatives: Math Quiz! Test

The derivative of -sin(x) is -cos(x).

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The given function is a quadratic function in the form of f(x) = x^2 - 3x + 2. To determine the nature of the extremum, we can find the derivative of the function, which is f'(x) = 2x - 3. Setting the derivative equal to zero and solving for x, we get x = 3/2.

Since the derivative is positive for x 3/2, we can conclude that the function has a minimum at x = 3/2. Therefore, the given answer "minimum at x = 3/2" is correct.

The correct answer is xx (1+ ln x). This can be determined by applying the product rule of differentiation. The derivative of xx is 1, and the derivative of ln x is 1/x. Therefore, using the product rule, the derivative of xx (1+ ln x) is xx (1/x) + 1 (xx) = xx + xx ln x = xx (1+ ln x).

The given expression is y = xsin x. To find y', we need to differentiate y with respect to x. Using the product rule, the derivative of xsin x is (sin x) + x(cos x). Therefore, the correct answer is x sin x [ (sin x)/x + ln x cos x].

The function f(x) = 2x^2 - 8x + 4 is a quadratic function with a positive leading coefficient (2), which means it opens upwards. When a quadratic function opens upwards, it is increasing on the interval where its derivative is positive. To find the derivative of f(x), we can differentiate the function with respect to x. The derivative of f(x) = 2x^2 - 8x + 4 is f'(x) = 4x - 8. Setting f'(x) > 0 and solving for x, we get x > 2. Therefore, the function f(x) is increasing in the interval [2, ∞].

The slope of a function at a given point can be found by taking the derivative of the function and evaluating it at that point. In this case, the function f(x) = x(2x-3)^(1/2) can be rewritten as f(x) = x√(2x-3). Taking the derivative of this function with respect to x using the product rule and chain rule, we get f'(x) = √(2x-3) + x(2x-3)^(-1/2)(2). Evaluating this derivative at x = 6, we get f'(6) = √(2(6)-3) + 6(2(6)-3)^(-1/2)(2) = 5. Therefore, the slope of the function at x = 6 is 5.

The given equation represents the positions of the two runners on the track. By setting the equation equal to zero and solving for t, we can find the time when they are crossing each other. In this case, when t = 2, the equation becomes 0 = 2(2-2), which simplifies to 0 = 0. This means that both runners are at the same position at t = 2 seconds, indicating that they are crossing each other at that time. Therefore, the correct answer is 2 sec.

The minimum attained velocity by the particle can be found by taking the derivative of the velocity function with respect to time and setting it equal to zero. By differentiating v = 2t^2 - 4t + 2, we get dv/dt = 4t - 4. Setting this equal to zero, we find t = 1. Substituting this value back into the velocity function, we get v = 2(1)^2 - 4(1) + 2 = 0 m/s. Therefore, the minimum attained velocity by the particle is 0 m/s.

The number of critical points or points of extremum can be determined by finding the derivative of the given function f(x) and then setting it equal to zero. In this case, the derivative of f(x) is f'(x) = x^3 + x^2 - 6x. By setting f'(x) = 0, we can solve for x and find the critical points. By solving the equation, we find that there are three critical points, which means the correct answer is 3.

The derivatives of the given functions are as follows:
1. The derivative of y = -6x^(1/6) with respect to x is -x^(-5/6).
2. The derivative of y = -sin(x) cot(x) with respect to x is -sin(x) - cos(x) cot(x).
3. The derivative of P = 4t with respect to t is 4.
4. The derivative of F = te^(-t) with respect to t is e^(-t) - te^(-t) (1 - t).

Therefore, the correct answer is -x^(-5/6), sin(x), 4t log(t), e^(-t) (1 - t).

The answer is 1 sec because it is the only option given that indicates a time at which they are running at the same speed. The other options do not provide any information about the speed of the individuals at a specific time. Therefore, the correct answer is 1 sec.