GATE DA 2025 | Question: 39

Question

Consider the function $f(x)=\dfrac{x^{3}}{3}+\dfrac{7}{2} x^{2}+10 x+\dfrac{133}{2}, x \in[-8,0]$. Which of the following statements is/are correct?

• A. The maximum value of $f$ is attained at $x=-5$
• B. The minimum value of $f$ is attained at $x=-2$
• C. The maximum value of $f$ is $\dfrac{133}{2}$
• D. The minimum value of the derivative of $f$ is attained at $x=-\dfrac{7}{2}$

Comments

Answer : C and D

Answer 1

So before going to answer we should know about one thing

When the domain is not given (for example, x∈(−∞,∞), the points we find by setting the derivative to zero are usually only local maxima or minima (highest or lowest just compared to nearby points).

But when a domain or interval is specified (like here x∈[−8,0] we are being asked for the global (absolute) maximum or minimum within that whole interval.

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f(x) = x^3/3 + (7/2)x^2 + 10x + 133/2 

1. First derivative:
f'(x) = 3x^2/3 + (7/2)*2x + 10 = x^2 + 7x + 10

Set f'(x)=0 so that we can find critical points

x^2 + 7x + 10 = 0
x^2 + 2x + 5x + 10 = x(x+2) + 5(x+2) = (x+5)(x+2) = 0
calculating we will get , x = -5 and x = -2

2. Second derivative:
f''(x) = d/dx (x^2 + 7x + 10) = 2x + 7 

3. Values:
f(-8) = 239/6 =39.83
f(-5) = 187/3 = 62.33 (local max)
f(-2) = 347/6 = 57.83 (local min)
f(0) = 133/2 = 66.5 (maximum) {we have calculated division so that we can know about absolute min and maximum}

So on [-8,0]:
Maximum value 66.5 at x=0
Local maximum at x=-5 (62.33)
Local minimum at x=-2 (57.83)

Absolute Maximum (Global Max) on [-8,0]

f(0) = 133/2

Absolute Minimum
f(-8) = 239/6

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Now come to statments:

A) Max value of f at x = -5
→ False, max is at x = 0 (as above)

B) Min value of f at x = -2
→ False, min is at x = -8 (as above)

C) Maximum value = 133/2
→ True, it occurs at f(0)

D) Minimum value of f'(x) is at x = -7/2 , so also true

f'(x) = x^2 + 7x + 10 (as calculated already)
f''(x) = 2x + 7 = 0
x = -7/2 {x value where f'(x) is smallest or largest}

Since f'(x) is a parabola, this must be minimum

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So options are C and D

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Comments

Thankyou my many doubts are cleared after this

Answer 2

1. Find the Derivative and Critical Points

The given function is:

Differentiating with respect to x:

Set (f'(x) = 0) to find the critical points:

Both critical points lie within the given interval [-8, 0].

2. Evaluate the Function at Critical Points and Endpoints

To determine the absolute maximum and minimum values, calculate (f(x)) at the critical points and boundary points:

• At (x = -8): f(-8) 
  • 
• At (x = -5) f(-5) (Local Maximum):
  • 
• At (x = -2) f(-2) (Local Minimum):
  • 
    
• At (x = 0) f(0) :
  • 
    

Evaluating the Statements

• Statement A: The maximum value of f is attained at(x = -5).
  • Incorrect. (x = -5) is only a local maximum. The global maximum occurs at (x = 0).
• Statement B: The minimum value of f is attained at (x = -2)
  • Incorrect. (x = -2) is only a local minimum. The global minimum occurs at (x = -8) 
• Statement C:The maximum value of f is 133/2.
  • Correct. The absolute maximum value on [-8, 0] is 66.5 = 133/2.
• Statement D: The minimum value of the derivative of f is attained at x = -7/2.
  • Correct. NOTE: Here, we are asked minimum value of derivative of f . Don't confuse it with minimum value of f .
  • WAY 1:The derivative function of   
  • For minimum value calculate critical points f''(x) = 2x + 7, x = -7/2 [ first derivative of (f'(x))
  • Check minimum: [f'''(x) = 2] > 0  (2nd derivative of (f'(x)) is >0 so it is local minimum)  
  • WAY 2:  is a parabola opening upwards. Its minimum occurs at the vertex: x = - (b/2a) = -(7/2(1)) = -7/2