The absolute minimum value of x^4-x^2-2x+5 is | vedclass

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(B) Let $f(x) = x^4-x^2-2x+5$. Find the derivative: $f'(x) = 4x^3-2x-2$. Set $f'(x) = 0$ to find critical points: $4x^3-2x-2 = 0 \Rightarrow 2x^3-x-1

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equal to $3$

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(B) Let $f(x) = x^4-x^2-2x+5$. Find the derivative: $f'(x) = 4x^3-2x-2$. Set $f'(x) = 0$ to find critical points: $4x^3-2x-2 = 0 \Rightarrow 2x^3-x-1 = 0$. Factoring the cubic equation: $2x^3-2x^2+2x^2-2x+x-1 = 0 \Rightarrow 2x^2(x-1) + 2x(x-1) + 1(x-1) = 0 \Rightarrow (x-1)(2x^2+2x+1) = 0$. The real root is $x = 1$. The quadratic factor $2x^2+2x+1$ has a negative discriminant $(D = 4-8 = -4)$,so it yields no real roots. Check the second derivative: $f''(x) = 12x^2-2$. At $x = 1$,$f''(1) = 12(1)^2-2 = 10 > 0$. Since $f''(1) > 0$,the function has a local minimum at $x = 1$. As $x 	o \pm \infty$,$f(x) 	o \infty$,so the local minimum is the absolute minimum. Calculating the value: $f(1) = (1)^4-(1)^2-2(1)+5 = 1-1-2+5 = 3$. Thus,the absolute minimum value is $3$.

The absolute minimum value of $x^4-x^2-2x+5$ is

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