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The function f(x) = x³ – 3x² – 9x – 7 has a stationary point at
To find the stationary points of the function f(x) = x³ - 3x² - 9x - 7, one would first have to compute the derivative of the function and set the function equal to zero. Stationary points occur at places where the derivative is equal to zero. First, differentiate f(x): f'(x) = d/dx(x³ - 3x² - 9x -Read more
To find the stationary points of the function f(x) = x³ – 3x² – 9x – 7, one would first have to compute the derivative of the function and set the function equal to zero. Stationary points occur at places where the derivative is equal to zero.
First, differentiate f(x):
f'(x) = d/dx(x³ – 3x² – 9x – 7) = 3x² – 6x – 9
Now, put f'(x) = 0 to get the stationary points:
3x² – 6x – 9 = 0
Divide the equation by 3:
x² – 2x – 3 = 0
Factor the quadratic equation:
(x – 3)(x + 1) = 0
Therefore, the solutions are x = 3 and x = -1.
Hence, the stationary points are at x = -1 and x = 3.
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The value of b for which the function f(x) = x + cos x + b is strictly decreasing over R is
In order to determine the value of b for which the function f(x) = x + cos x + b is strictly decreasing over ℝ, we start by analyzing the derivative of the function. Let us differentiate f(x): f'(x) = d/dx(x + cos x + b) = 1 - sin x Hence for being strictly decreasing, the derivative must be negativRead more
In order to determine the value of b for which the function f(x) = x + cos x + b is strictly decreasing over ℝ, we start by analyzing the derivative of the function. Let us differentiate f(x):
f'(x) = d/dx(x + cos x + b) = 1 – sin x
Hence for being strictly decreasing, the derivative must be negative for all values of x; i.e., f'(x) < 0.
1 – sin x 1
But the function sin x can never be more than 1 for any real number x. Thus, b cannot be taken such that for all real number x, f'<0.
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The least value of the function f(x) = 2cos x + x in the closed interval [0, π/2] is
We find the minimum value of the function f(x) = 2 cos x + x over the closed interval [0, π/2] by determining its critical numbers and then we check the value of the function at the interval's endpoints. Compute the derivative of the function: f'(x) = d/dx(2 cos x + x) = -2 sin x + 1 Set the derivatRead more
We find the minimum value of the function f(x) = 2 cos x + x over the closed interval [0, π/2] by determining its critical numbers and then we check the value of the function at the interval’s endpoints.
Compute the derivative of the function:
f'(x) = d/dx(2 cos x + x) = -2 sin x + 1
Set the derivative equal to zero to find critical points:
-2 sin x + 1 = 0
sin x = 1/2
The solution to sin x = 1/2 in the interval [0, π/2] is x = π/6.
Evaluate f(x) at the critical point and the endpoints of the interval:
– At x = 0:
f(0) = 2 cos(0) + 0 = 2 × 1 + 0 = 2
– At x = π/6:
f(π/6) = 2 cos(π/6) + π/6 = 2 × (√3/2) + π/6 = √3 + π/6
– At x = π/2:
f(π/2) = 2 cos(π/2) + π/2 = 2 × 0 + π/2 = π/2
Compare the values:
– f(0) = 2
– f(π/6) = √3 + π/6 ≈ 1.732 + 0.523 = 2.255
– f(π/2) = π/2 ≈ 1.570
The minimum value of the function in the interval [0, π/2] is attained at f(0) = 2.
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Find the intervals in which the function f given by f(x) = x² – 4x + 6 is strictly increasing:
To find the intervals where the function f(x) = x² - 4x + 6 is strictly increasing, we have to look at its derivative. Find the derivative of the function: f'(x) = d/dx(x² - 4x + 6) = 2x - 4 Determine where the derivative is positive (for strictly increasing behavior): f'(x) > 0 2x - 4 > 0 xRead more
To find the intervals where the function f(x) = x² – 4x + 6 is strictly increasing, we have to look at its derivative.
Find the derivative of the function:
f'(x) = d/dx(x² – 4x + 6) = 2x – 4
Determine where the derivative is positive (for strictly increasing behavior):
f'(x) > 0
2x – 4 > 0
x > 2
So, the function is strictly increasing when x > 2.
Conclusion:
The function is strictly increasing on the interval (2, ∞).
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The interval, in which function y = x³ + 6x² + 6 is increasing is
To find the interval for which the function y = x³ + 6x² + 6 is increasing, we will look at its derivative. Compute the derivative of the function: y'(x) = d/dx(x³ + 6x² + 6) = 3x² + 12x Determine where the derivative is positive, so the function is increasing: y'(x) > 0 3x² + 12x > 0 Factor tRead more
To find the interval for which the function y = x³ + 6x² + 6 is increasing, we will look at its derivative.
Compute the derivative of the function:
y'(x) = d/dx(x³ + 6x² + 6) = 3x² + 12x
Determine where the derivative is positive, so the function is increasing:
y'(x) > 0
3x² + 12x > 0
Factor the expression:
3x(x + 4) > 0
This inequality holds when x 0. Thus, the function is increasing in the intervals (-∞, -4) and (0, ∞).
Conclusion:
The correct intervals where the function is increasing are (-∞, 0) U (4, ∞).
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