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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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If A is a matrix of order 3 x 3 such that A² = 4A – 3I, then A⁻¹ is
We are given a 3×3 matrix A that satisfies the equation: A² = 4A - 3I Step 1: Expressing A⁻¹ To find A⁻¹, we rearrange the given equation: A² - 4A + 3I = 0 Factoring, (A - I)(A - 3I) = 0 This implies that A satisfies the equation: (A - I)(A - 3I) = 0 Multiplying both sides by (A - I)⁻¹ (if it existsRead more
We are given a 3×3 matrix A that satisfies the equation:
A² = 4A – 3I
Step 1: Expressing A⁻¹
To find A⁻¹, we rearrange the given equation:
A² – 4A + 3I = 0
Factoring,
(A – I)(A – 3I) = 0
This implies that A satisfies the equation:
(A – I)(A – 3I) = 0
Multiplying both sides by (A – I)⁻¹ (if it exists),
A – 3I = (A – I)⁻¹ 0
Since A – I is invertible, we take its inverse on both sides,
A⁻¹ = 1/3 (4I – A)
Step 2: Selecting the Correct Option
Comparing with the given choices, the correct answer is: 1/3 (4I – A)
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If a matrix has 5 elements, write all possible orders it can have are
A matrix with 5 elements can have different possible orders, provided that the total number of elements is the product of its rows and columns. Let the number of rows be m and the number of columns be n, then: m × n = 5 The possible values of (m, n) that satisfy this equation are: - 1 × 5 (1 row andRead more
A matrix with 5 elements can have different possible orders, provided that the total number of elements is the product of its rows and columns.
Let the number of rows be m and the number of columns be n, then:
m × n = 5
The possible values of (m, n) that satisfy this equation are:
– 1 × 5 (1 row and 5 columns)
– 5 × 1 (5 rows and 1 column)
Since both 1 × 5 and 5 × 1 are possible, the correct answer is:
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