Application of Derivatives Exercise 02
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:

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.

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

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).

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).

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

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.

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).

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:
 First, we need to find the derivative of the function f(x) = x^2 + ax + 1.
f’(x) = 2x + a
 We then need to set the derivative equal to 0 and solve for a.
2x + a = 0 a = 2x
 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
 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:

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).

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

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).

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).

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

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).

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

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).

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:

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

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

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

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

f’(x)=0 when x=2.

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.

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

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

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:

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.

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).

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=∞.

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).

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.

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).

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.

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^(n1).
Step 4: In the case of f(x)=3x+17, a=3 and n=1, so the derivative of the function is 3*1^(11) = 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:
 Find the first derivative of the function: f’(x)=6x^26x36
 Set the first derivative equal to zero and solve for x: 6x^26x36=0; x=(3±√33)/6
 Find the second derivative of the function: f’’(x)=12x6
 Determine the sign of the second derivative at each xvalue: f’’(3±√33)/6 = 12(3±√33)/6  6 = ±√33
 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:
 Find the first derivative of the function: f’(x)=6x^26x36
 Set the first derivative equal to zero and solve for x: 6x^26x36=0; x=(3±√33)/6
 Find the second derivative of the function: f’’(x)=12x6
 Determine the sign of the second derivative at each xvalue: f’’(3±√33)/6 = 12(3±√33)/6  6 = ±√33
 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(x2)^2 + (x)(2)(x2)
Step 2: Set y’ = 0 and solve for x.
2x(x2)^2 + (x)(2)(x2) = 0
2x(x2)(x2) + 2x(x2) = 0
2x(x2)(x2 + 1) = 0
2x(x2)(x1) = 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(x2)]^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:

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

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

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

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

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

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

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