Applications Of Derivatives - II
ICSE · Class 12 · Mathematics
Flashcards for Applications Of Derivatives - II — ICSE Class 12 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.
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![Absolute (Global) Maxima and Minima A graph of a continuous function on a closed interval [a, b], clearly indicating the absolute maximum and absolute minimum values and their corresponding points.](https://s3.ap-southeast-2.amazonaws.com/super-tutor/production/educational-images/mathematics/mathematics_global_maxima_minima_graph.png)
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Get startedSolve for the absolute maximum and absolute minimum of f(x)=|x| on R.
Answer
Step 1: For every real x, |x| >= 0. Step 2: The smallest value is 0, and it occurs at x = 0. Step 3: There is no largest value because |x| keeps growing as |x| becomes larger. Answer: Absolute minimum…
Find the absolute maximum and absolute minimum of f(x)=|x| on [2,5].
Answer
Step 1: On [2,5], x is positive, so |x| = x. Step 2: The smallest value on the interval is at x = 2. Step 3: The largest value on the interval is at x = 5. Step 4: f(2)=2 and f(5)=5. Answer: Absolute …
Why can a local maximum or local minimum not occur at an end point of an interval?
Answer
A local extremum needs a whole neighbourhood around the point. At an end point, there is no open interval on both sides inside the domain, so the neighbourhood condition fails. Example: an end point c…
Find the critical points of f(x)=x^3+x^2+x+1.
Answer
Step 1: Differentiate: f'(x)=3x^2+2x+1. Step 2: Set f'(x)=0: 3x^2+2x+1=0. Step 3: The discriminant is 2^2-4(3)(1)=4-12=-8, so there are no real roots. Step 4: f'(x) does not fail to exist anywhere. An…
Solve: Find the local maximum or minimum of f(x)=2x^3-6x^2+6x+5 using the first derivative test.
Answer
Step 1: Differentiate: f'(x)=6x^2-12x+6=6(x-1)^2. Step 2: Critical point: f'(x)=0 gives x=1. Step 3: Check sign near x=1. Since 6(x-1)^2 is positive on both sides of 1, the sign does not change. Step …
Find the local maximum and local minimum of f(x)=(x-3)^4 using the first derivative test.
Answer
Step 1: f'(x)=4(x-3)^3. Step 2: Set f'(x)=0 gives x=3. Step 3: For x slightly less than 3, f'(x)<0. For x slightly more than 3, f'(x)>0. Step 4: Negative to positive means local minimum. Step 5: f(3)=…
Find the local maximum of f(x)=x*sqrt(1-x) on (0,1).
Answer
Step 1: Differentiate. f'(x)=x*(-1/(2sqrt(1-x)))+sqrt(1-x)=(2-3x)/(2sqrt(1-x)). Step 2: Set numerator equal to zero: 2-3x=0, so x=2/3. Step 3: For x slightly less than 2/3, f'(x)>0. Step 4: For x slig…
Use the second derivative test for f(x)=2x^3-21x^2+36x-20.
Answer
Step 1: f'(x)=6x^2-42x+36=6(x-1)(x-6). Step 2: Critical points are x=1 and x=6. Step 3: f''(x)=12x-42. Step 4: f''(1)= -30 < 0, so x=1 is a local maximum. Step 5: f(1)=2-21+36-20=-3. Step 6: f''(6)=30…
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