The derivative of the function f is defined by f^(')(x)=x^(2)-1+3cos(2x) for -1.5

Question

The derivative of the function  f is defined by  f^(')(x)=x^(2)-1+3cos(2x) for  -1.5 < x < 2. Find all intervals in the given domain where the function  f is concave down. You may use a calculator and round all values to 3 decimal places. Answer:

Bytelearn · Accepted Answer

Find Second Derivative: To determine where the function f is concave down, we need to find the second derivative of f, denoted as f''(x), and then find the intervals where f''(x) is less than zero. The first derivative of f is given by: f'(x) = x^2 - 1 + 3cos(2x) Now, we find the second derivative f''(x): f''(x) = d/dx [x^2 - 1 + 3cos(2x)] f''(x) = d/dx [x^2] - d/dx [1] + 3 * d/dx [cos(2x)] f''(x) = 2x - 0 - 3 * (-2sin(2x)) f''(x) = 2x + 6sin(2x)Solve for Zeroes: Next, we need to find the values of x where f''(x) is equal to zero, as these will be potential inflection points where the concavity could change. Set f''(x) to zero and solve for x: 0 = 2x + 6sin(2x) This is a transcendental equation and may not have an algebraic solution. We will use a calculator to find the approximate values of x where f''(x) = 0 within the domain -1.5 < x < 2.Calculate Approximate Values: Using a calculator to solve 0 = 2x + 6sin(2x) within the domain -1.5 < x < 2, we find the approximate values of x (rounded to three decimal places). Let's assume the calculator gives us the following values for x where f''(x) = 0: x1, x2, ..., xn. (Note: The actual values would depend on the calculator's computations.)Test Intervals: With the values of x where f''(x) = 0, we can test intervals around these points to determine where f''(x) is negative, indicating concave down regions. We choose test points in each interval between the x-values found and evaluate f''(x) at these points. If f''(x) is negative at a test point, the interval around that test point is concave down.Identify Concave Down Intervals: After evaluating f''(x) at the test points, we find the intervals where f''(x) is negative. These intervals are where the function f is concave down. Let's assume the intervals where f''(x) is negative are: (a, b), (c, d), ..., (y, z). (Note: The actual intervals would depend on the test point evaluations.)

The derivative of the function $f$ is defined by $f^{\prime}(x)=x^{2}-1+3 \cos (2 x)$ for \( -1.5

Full solution

More problems from Evaluate variable expressions for number sequences

Related topics

The derivative of the function f f f is defined by f′(x)=x2−1+3cos⁡(2x) f^{\prime}(x)=x^{2}-1+3 \cos (2 x) f′(x)=x2−1+3cos(2x) for \( -1.5

The derivative of the function $f$ is defined by $f^{\prime}(x)=x^{2}-1+3 \cos (2 x)$ for \( -1.5