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Example 29 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at March 16, 2023 by Teachoo
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Example 29 You are here
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Chapter 6 Class 12 Application of Derivatives
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Transcript
Example 29 (Method 1) Find all points of local maxima and local minima of the function f given by π (π₯)=π₯3 β 3π₯ + 3. π(π₯)=π₯3 β 3π₯+3 Finding π^β² (π) πβ²(π₯)= 3π₯^2 β 3+0 πβ²(π₯)= 3(π₯^2β1) Putting πβ²(π)= 0 3(π₯^2β1)=0 π₯^2β1=0 (π₯β1)(π₯+1)=0 Thus π = 1 , β1 are only critical points Finding maximum & minimum value of π(π₯)=π₯3 β 3π₯+3 Minimum value π(1)=(1)3 β3(1)+3= 1 β 3 + 3 = 1 Maximum value π(β1)=(β1)3 β3(β1)+3= β1 +3 + 3 = 5 Example 29 (Method 2) Find all points of local maxima and local minima of the function f given by π (π₯) = π₯3 β 3π₯ + 3. π(π₯)=π₯3 β 3π₯+3 Finding π^β² (π) πβ²(π₯)= 3π₯^2 β 3+0 πβ²(π₯)= 3(π₯^2β1) Putting πβ²(π)= 0 3(π₯^2β1)=0 π₯^2β1=0 (π₯β1)(π₯+1)=0 So, x = 1 & x = β1 Finding πβ²β²(π) π^β² (π₯)=3(π₯^2β1) π^β²β² (π₯)=3 π(π₯^2 β 1)/ππ₯ π^β²β² (π₯)=3(2π₯β0) π^β²β² (π₯)=6π₯ Putting π=βπ π^β²β² (π₯) = 6(β1) <0 π₯ = β1 is point of local maxima Putting π=π π^β²β² (π₯) = 6(1) >0 π₯ = 1 is point of local minima Finding maximum & minimum value of π(π₯)=π₯3 β 3π₯+3 Minimum value at x = 1 π(1)=(1)3 β3(1)+3= 1 β 3 + 3 = 1 Maximum value at x = β1 π(β1)=(β1)3 β3(β1)+3= β1 + 3 + 3 = 5
