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Exercise 7.7 · Q3

Q.For the function f(x)=4x3+3x2−6x+1f(x)=4x^3+3x^2-6x+1 find the intervals of monotonicity, local extrema, intervals of concavity and points of inflection.

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Compute f′f' for monotonicity/extrema (first derivative test) and f′′f'' for concavity/inflection, in the usual order.

Step 1. Differentiate for f′f' and factor.

f(x)=4x3+3x2−6x+1⇒f′(x)=12x2+6x−6=6(2x2+x−1)=6(2x−1)(x+1)f(x)=4x^3+3x^2-6x+1\Rightarrow f'(x)=12x^2+6x-6=6(2x^2+x-1)=6(2x-1)(x+1).

Critical numbers: x=−1, 12x=-1,\ \dfrac12.

Step 2. Sign table for f′f'.

x<−1x<-1: (2x−1)<0,(x+1)<0⇒f′>0(2x-1)<0,(x+1)<0\Rightarrow f'>0 (increasing).

−1<x<12-1<x<\tfrac12: (2x−1)<0,(x+1)>0⇒f′<0(2x-1)<0,(x+1)>0\Rightarrow f'<0 (decreasing).

x>12x>\tfrac12: both >0⇒f′>0>0\Rightarrow f'>0 (increasing).

Local max at x=−1x=-1: f(−1)=−4+3+6+1=6f(-1)=-4+3+6+1=6. Local min at x=12x=\tfrac12: f ⁣(12)=4 ⁣(18)+3 ⁣(14)−3+1=12+34−3+1=−34f\!\left(\tfrac12\right)=4\!\left(\tfrac18\right)+3\!\left(\tfrac14\right)-3+1=\tfrac12+\tfrac34-3+1=-\tfrac34.

Step 3. Differentiate again for f′′f''.

f′′(x)=24x+6=6(4x+1)f''(x)=24x+6=6(4x+1). Zero at x=−14x=-\dfrac14.

For x<−14x<-\tfrac14: f′′<0f''<0 (concave down). For x>−14x>-\tfrac14: f′′>0f''>0 (concave up) — a genuine sign change, so a point of inflection.

Step 4. Inflection point value. …

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