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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

equal to $3$

Vedclass · Answer

(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

Similar Questions

If $A=\{x : 9x \geq x^2+20\}$ and $f: A \rightarrow R$ is defined by $f(x)=2x^3-15x^2+36x-48$,then the maximum value of $f(x)$ is

The least value of $a$ for which the equation $\frac{4}{\sin x} + \frac{1}{1 - \sin x} = a$ has at least one solution on the interval $(0, \pi/2)$ is:

The minimum value of the function $f(x) = x^{10} + x^2 + \frac{1}{x^{12}} + \frac{1}{1 + \sec^{-1} x}$ is

The local maximum value of the function $f(x)=-(x-2)^3(x+2)^2$ is

At what points in the interval $[0, 2\pi]$,does the function $\sin 2x$ attain its maximum value?

Vedclass Test Series

Exam Paper Generator

Online Exam Module