Question:

Prove that the function f given by f(x)=logsinx is strictly increasing on (0,π​/2) and strictly decreasing on (π​/2,π)

Answer:

1. To prove that a function is strictly increasing on a given interval, we must show that the derivative of the function is greater than zero on that interval.

2. The derivative of f(x) is f’(x)=cosx/sinx.

3. To show that f’(x) is greater than zero on (0,π/2), we must show that cosx is greater than zero on (0,π/2).

4. Since the cosine function is positive on the interval (0,π/2), we can conclude that f’(x) is greater than zero on (0,π/2).

5. Therefore, the function f(x)=logsinx is strictly increasing on (0,π/2).

6. To show that the function f(x) is strictly decreasing on (π/2,π), we must show that the derivative of the function is less than zero on that interval.

7. Since the cosine function is negative on the interval (π/2,π), we can conclude that f’(x) is less than zero on (π/2,π).

8. Therefore, the function f(x)=logsinx is strictly decreasing on (π/2,π).

Question:

Prove that the logarithmic function is strictly increasing on (0,∞).

Answer:

Proof:

Step 1: Assume f(x) = log(x), where x ∈ (0,∞).

Step 2: Show that f’(x) > 0 for all x ∈ (0,∞).

Step 3: Since f’(x) > 0 for all x ∈ (0,∞), this implies that the function f(x) is strictly increasing on (0,∞).

Step 4: Therefore, the logarithmic function is strictly increasing on (0,∞).

Question:

Which of the following functions are strictly decreasing on (0,π​/2)? This question has multiple correct options A cosx B cos2x C cos3x D tanx

Answer:

Step 1: Determine the domain of the functions. The domain of all the functions is (0,π/2).

Step 2: Determine the range of each function. The range of A (cosx) is [-1,1], the range of B (cos2x) is [-1,1], the range of C (cos3x) is [-1,1], and the range of D (tanx) is (-∞,∞).

Step 3: Determine if the function is strictly decreasing on the given domain. A (cosx) is not strictly decreasing on (0,π/2). B (cos2x) is not strictly decreasing on (0,π/2). C (cos3x) is strictly decreasing on (0,π/2). D (tanx) is strictly decreasing on (0,π/2).

Therefore, the correct answer is C (cos3x) and D (tanx).

Question:

Find the least value of a such that the function f given by f(x)=x^2+ax+1 is strictly increasing on [1,2].

Answer:

1. First, we need to find the derivative of the function f(x) = x^2 + ax + 1.

f’(x) = 2x + a

2. We then need to set the derivative equal to 0 and solve for a.

2x + a = 0 a = -2x

3. Since we want to find the least value of a such that the function is strictly increasing on [1,2], we need to plug in the two endpoints of the interval and solve for a.

When x = 1, a = -2

When x = 2, a = -4

4. Therefore, the least value of a such that the function is strictly increasing on [1,2] is a = -4.

Question:

Prove that the function f given by f(x)=logcosx is strictly decreasing on (0,π​/2)and strictly increasing on (π​/2,π).

Answer:

1. To prove that f is strictly decreasing on (0,π​/2) we need to show that the derivative of f is negative for all x in (0,π​/2).

2. The derivative of f is f’(x) = -sin(x)/cos(x).

3. To show that f’(x) is negative for all x in (0,π​/2), we need to show that -sin(x) is negative for all x in (0,π​/2).

4. Since sin(x) is negative for all x in (0,π​/2), it follows that -sin(x) is negative for all x in (0,π​/2).

5. Therefore, f’(x) is negative for all x in (0,π​/2), which proves that f is strictly decreasing on (0,π​/2).

6. To prove that f is strictly increasing on (π​/2,π), we need to show that the derivative of f is positive for all x in (π​/2,π).

7. Since f’(x) = -sin(x)/cos(x), we need to show that -sin(x) is positive for all x in (π​/2,π).

8. Since sin(x) is positive for all x in (π​/2,π), it follows that -sin(x) is positive for all x in (π​/2,π).

9. Therefore, f’(x) is positive for all x in (π​/2,π), which proves that f is strictly increasing on (π​/2,π).

Question:

Prove that the function given by f(x)=x^3−3x^2+3x−100 is increasing in R.

Answer:

1. To prove that a function is increasing, we must show that the derivative of the function is always positive.

2. The derivative of f(x) is given by f’(x)=3x^2−6x+3.

3. We must show that f’(x) is always greater than zero for all x in the real numbers.

4. To do this, we must set f’(x) equal to zero and solve for x.

5. f’(x)=0 when x=2.

6. We must now check if f’(x) is positive for all values of x less than 2 and all values of x greater than 2.

7. For x<2, f’(x)=3x^2−6x+3>3(2)^2−6(2)+3=3>0.

8. For x>2, f’(x)=3x^2−6x+3>3(2)^2−6(2)+3=3>0.

9. Since f’(x) is always positive for all x in the real numbers, we can conclude that the function f(x)=x^3−3x^2+3x−100 is increasing in R.

Question:

Show that the function given by f(x)=e^2x is strictly increasing on R.

Answer:

Step 1: Recall that a function is strictly increasing if for all x1, x2 in the domain of f, x1 < x2 implies that f(x1) < f(x2).

Step 2: Let x1 and x2 be any two real numbers such that x1 < x2.

Step 3: Then, f(x1) = e^2x1 and f(x2) = e^2x2.

Step 4: Since e^2x1 < e^2x2 (since e is always positive and 2x1 < 2x2), we can conclude that f(x1) < f(x2), which means that the function f(x) = e^2x is strictly increasing on R.

Question:

Prove that the function f given by f(x)=x^2−x+1 is neither strictly increasing nor strictly decreasing on (−1,1).

Answer:

Step 1: Understand the problem. The problem is asking us to prove that the function f(x)=x^2−x+1 is neither strictly increasing nor strictly decreasing on the interval (−1,1).

Step 2: Analyze the function. f’(x)=2x−1

Step 3: Find the critical points. The critical points are the points where the derivative of the function is equal to zero.

2x−1=0 x=1/2

Step 4: Analyze the sign of the derivative at the critical points. At x=1/2, the derivative is negative (f’(1/2)=-1/2).

Step 5: Determine if the function is increasing or decreasing. Since the derivative is negative at the critical point, the function is decreasing on the left side of the critical point (x<1/2) and increasing on the right side (x>1/2). Therefore, the function is neither strictly increasing nor strictly decreasing on (−1,1).

Question:

Find the intervals in which the following functions are strictly increasing or decreasing: (a) x^2+2x−5 (b) 10−6x−2x^2 (c) −2x^3−9x^2−12x+1 (d) 6−9x−x^2 (e) f(x)=(x+1)^3(x−3)^3

Answer:

(a) The function x^2+2x−5 is strictly decreasing on the interval (-∞, -1) and strictly increasing on the interval (-1, ∞).

(b) The function 10−6x−2x^2 is strictly decreasing on the interval (-∞, 5/3) and strictly increasing on the interval (5/3, ∞).

(c) The function −2x^3−9x^2−12x+1 is strictly decreasing on the interval (-∞, -2) and strictly increasing on the interval (-2, ∞).

(d) The function 6−9x−x^2 is strictly decreasing on the interval (-∞, 3) and strictly increasing on the interval (3, ∞).

(e) The function f(x)=(x+1)^3(x−3)^3 is strictly increasing on the interval (-∞, -1) and (-3, ∞).

Question:

The interval in which y=x^2e^x is decreasing is. A (−∞,∞) B (2,0) C (2,∞) D (0,2)

Answer:

Answer: B (2,0)

Question:

Prove that the function f given by f(x)=logcosx is strictly decreasing on (0,π​/2)and strictly increasing on (π​/2,π).

Answer:

1. To prove that the function f(x)=logcosx is strictly decreasing on (0,π/2), we must show that the derivative of f(x) is negative for all x in that interval.

2. To find the derivative of f(x), we must use the chain rule. The derivative of f(x) is f’(x)= -sin(x)log(cos(x))*sec2(x).

3. We can now substitute in x=0 and x=π/2 to determine the sign of the derivative. When x=0, f’(x)=-sin(0)log(cos(0))sec2(0)=-1log(1)*1=-1. When x=π/2, f’(x)=-sin(π/2)log(cos(π/2))sec2(π/2)=-1log(0)*1=-∞.

4. Since the derivative is negative at both x=0 and x=π/2, it is negative over the entire interval (0,π/2). Therefore, f(x) is strictly decreasing on (0,π/2).

5. To prove that the function f(x)=logcosx is strictly increasing on (π/2,π), we must show that the derivative of f(x) is positive for all x in that interval.

6. We can again use the chain rule to find the derivative of f(x). The derivative of f(x) is f’(x)= -sin(x)log(cos(x))*sec2(x).

7. We can now substitute in x=π/2 and x=π to determine the sign of the derivative. When x=π/2, f’(x)=-sin(π/2)log(cos(π/2))sec2(π/2)=-1log(0)*1=-∞. When x=π, f’(x)=-sin(π)log(cos(π))sec2(π)=1log(1)*1=1.

8. Since the derivative is positive at both x=π/2 and x=π, it is positive over the entire interval (π/2,π). Therefore, f(x) is strictly increasing on (π/2,π).

Question:

Show that the function given by f(x)=3x+17 is strictly increasing on R.

Answer:

Answer: Step 1: To show that a function is strictly increasing, we need to show that the derivative of the function is greater than 0.

Step 2: To find the derivative of the function f(x)=3x+17, we use the power rule for derivatives.

Step 3: The power rule states that the derivative of a function of the form f(x)=ax^n is n*ax^(n-1).

Step 4: In the case of f(x)=3x+17, a=3 and n=1, so the derivative of the function is 3*1^(1-1) = 3.

Step 5: Since the derivative of the function is greater than 0, the function is strictly increasing on R.

Question:

Find the intervals in which the function f given by f(x)=2x^3−3x^2−36x+7 is (a) strictly increasing (b) strictly decreasing.

Answer:

(a) Strictly Increasing:

1. Find the first derivative of the function: f’(x)=6x^2-6x-36
2. Set the first derivative equal to zero and solve for x: 6x^2-6x-36=0; x=(3±√33)/6
3. Find the second derivative of the function: f’’(x)=12x-6
4. Determine the sign of the second derivative at each x-value: f’’(3±√33)/6 = 12(3±√33)/6 - 6 = ±√33
5. Determine the intervals in which the function is strictly increasing: f is strictly increasing when x < 3-√33 and x > 3+√33

(b) Strictly Decreasing:

1. Find the first derivative of the function: f’(x)=6x^2-6x-36
2. Set the first derivative equal to zero and solve for x: 6x^2-6x-36=0; x=(3±√33)/6
3. Find the second derivative of the function: f’’(x)=12x-6
4. Determine the sign of the second derivative at each x-value: f’’(3±√33)/6 = 12(3±√33)/6 - 6 = ±√33
5. Determine the intervals in which the function is strictly decreasing: f is strictly decreasing when x > 3-√33 and x < 3+√33

Question:

Let I be any interval disjoint from [−1,1]. Prove that the function f given byf(x)=x+1​/x is strictly increasing in interval disjoint from I.

Answer:

Step 1: First, we need to define the function f(x). We can write it as f(x) = x + 1/x.

Step 2: Next, we need to prove that the function is strictly increasing in the interval I. To do this, we will use the definition of a strictly increasing function. A function is strictly increasing if and only if for all x and y in the domain of the function, if x < y then f(x) < f(y).

Step 3: To prove that f is strictly increasing in the interval I, we need to show that for all x and y in I, if x < y then f(x) < f(y).

Step 4: To prove this, we will use the definition of a strictly increasing function and prove the statement by contradiction. Suppose that for some x and y in I, if x < y then f(x) ≥ f(y).

Step 5: We can rearrange the inequality to get f(x) - f(y) ≥ 0.

Step 6: We can then rewrite the equation as (x + 1/x) - (y + 1/y) ≥ 0.

Step 7: We can then simplify the equation to get x - y + 1/x - 1/y ≥ 0.

Step 8: We can then rearrange the equation to get x - y + (1/y - 1/x) ≥ 0.

Step 9: Since x and y are in I, which is disjoint from [-1,1], we know that x and y are not equal to -1 or 1. Therefore, 1/y - 1/x > 0.

Step 10: Since 1/y - 1/x > 0, we can rewrite the equation as x - y > 0.

Step 11: Since x - y > 0, this contradicts our assumption that x < y. Therefore, our assumption is false and f(x) < f(y).

Step 12: Since we have shown that for all x and y in I, if x < y then f(x) < f(y), this proves that the function f is strictly increasing in the interval I.

Question:

Show that the function given by f(x)=sinx is (a) strictly increasing in (0,π​/2) (b) strictly decreasing in (π​/2,π) (c) neither increasing nor decreasing in (0,π).

Answer:

(a) To show that the function f(x)=sin x is strictly increasing in (0,π/2):

Step 1: Calculate the derivative of f(x)=sin x.

f’(x)=cos x

Step 2: Show that the derivative of f(x)=sin x is greater than 0 in the interval (0,π/2).

cos x > 0 for x ∈ (0,π/2).

Step 3: Since the derivative of f(x)=sin x is greater than 0 in the interval (0,π/2), it is strictly increasing in this interval.

(b) To show that the function f(x)=sin x is strictly decreasing in (π/2,π):

Step 1: Calculate the derivative of f(x)=sin x.

f’(x)=cos x

Step 2: Show that the derivative of f(x)=sin x is less than 0 in the interval (π/2,π).

cos x < 0 for x ∈ (π/2,π).

Step 3: Since the derivative of f(x)=sin x is less than 0 in the interval (π/2,π), it is strictly decreasing in this interval.

(c) To show that the function f(x)=sin x is neither increasing nor decreasing in (0,π):

Step 1: Calculate the derivative of f(x)=sin x.

f’(x)=cos x

Step 2: Show that the derivative of f(x)=sin x is neither greater than 0 nor less than 0 in the interval (0,π).

cos x is neither greater than 0 nor less than 0 for x ∈ (0,π).

Step 3: Since the derivative of f(x)=sin x is neither greater than 0 nor less than 0 in the interval (0,π), it is neither increasing nor decreasing in this interval.

Question:

Find the values of x for which y=[x(x−2)]^2 is an increasing function.

Answer:

Step 1: Determine the derivative of y with respect to x.

y’ = 2x(x-2)^2 + (x)(2)(x-2)

Step 2: Set y’ = 0 and solve for x.

2x(x-2)^2 + (x)(2)(x-2) = 0

2x(x-2)(x-2) + 2x(x-2) = 0

2x(x-2)(x-2 + 1) = 0

2x(x-2)(x-1) = 0

x = 0, x = 2, x = 1

Step 3: Plug each of the solutions from Step 2 into the original equation to determine whether the equation is increasing or decreasing.

x = 0: y = 0 (constant)

x = 2: y = 0 (constant)

x = 1: y = 1 (increasing)

Therefore, the only value of x for which y = [x(x-2)]^2 is an increasing function is x = 1.

Question:

Find the intervals in which the function f given by f(x)=2x^2−3x is (a) strictly increasing (b) strictly decreasing.

Answer:

(a) Strictly Increasing:

1. Find the first derivative of the function f(x) = 2x^2 - 3x f’(x) = 4x - 3

2. Set the first derivative equal to zero and solve for x: 4x - 3 = 0 x = 3/4

3. Find the second derivative of the function f(x) = 2x^2 - 3x f’’(x) = 4

4. Since the second derivative is positive, the function is strictly increasing on the interval (-∞, 3/4) and (3/4, ∞).

(b) Strictly Decreasing:

1. Find the first derivative of the function f(x) = 2x^2 - 3x f’(x) = 4x - 3

2. Set the first derivative equal to zero and solve for x: 4x - 3 = 0 x = 3/4

3. Find the second derivative of the function f(x) = 2x^2 - 3x f’’(x) = 4

4. Since the second derivative is negative, the function is strictly decreasing on the interval (-∞, 3/4) and (3/4, ∞).

Question:

On which of the following intervals is the function f given by f(x)=x^100+sinx−1 strictly decreasing ? A (0,1) B (π​/2,π) C (0,π​/2) D None of these

Answer:

Step 1: Determine the derivative of the function f(x).

f’(x) = 100x^99 + cosx

Step 2: Determine the sign of the derivative on each of the given intervals.

A (0,1): The derivative is positive at x = 0 and negative at x = 1, so the function is not strictly decreasing.

B (π/2,π): The derivative is negative at x = π/2 and positive at x = π, so the function is not strictly decreasing.

C (0,π/2): The derivative is positive at x = 0 and negative at x = π/2, so the function is strictly decreasing.

Step 3: Answer the question.

The function f is strictly decreasing on the interval (0,π/2).