Question

Find the local manimum f(x)=x3-3x

Answer 1

$f(x) = {x}^{3} - 3x \\ \\ f'(x) = \frac{d}{dx} \bigg( {x}^{3} - 3x \bigg) = 3 {x}^{2} - 3 \\ \\ f'(x)=0\\ \\ \Rightarrow 3 {x}^{2} - 3 =0\\ \\ \Rightarrow 3 {x}^{2} =3\\ \\ \Rightarrow x^2 = 1\\ \\ \Rightarrow x =\pm 1\\ \\ \\ f"(x) = \frac{d}{dx} \bigg( 3 {x}^{2} - 3 \bigg) = 6x\\ \\ f"(1)= 6 \times 1 = 6 > 0\\ \text{at x = 1, it attains minima}\\ minima=f(1)=1^3 - 3 \times 1 = 1-1 = 0\\ \\ f"(-1)= 6 \times (-1) = -6 < 0\\ \text{at x = -1, it attains maxima}\\ maxima=f(-1)=(-1)^3-3 \times (-1) = -1+3=2$

Answer 2

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$f(x) = {x}^{3} - 3x \\ \\ f'(x) = \frac{d}{dx} \bigg( {x}^{3} - 3x \bigg) = 3 {x}^{2} - 3 \\ \\ f'(x)=0\\ \\ \Rightarrow 3 {x}^{2} - 3 =0\\ \\ \Rightarrow 3 {x}^{2} =3\\ \\ \Rightarrow x^2 = 1\\ \\ \Rightarrow x =\pm 1\\ \\ \\ f"(x) = \frac{d}{dx} \bigg( 3 {x}^{2} - 3 \bigg) = 6x\\ \\ f"(1)= 6 \times 1 = 6 > 0\\ \text{at x = 1, it attains minima}\\ minima=f(1)=1^3 - 3 \times 1 = 1-1 = 0\\ \\ f"(-1)= 6 \times (-1) = -6 < 0\\ \text{at x = -1, it attains maxima}\\ maxima=f(-1)=(-1)^3-3 \times (-1) = -1+3=2$

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