Let f: R rightarrow R be given by f(x) = |x^2 | vedclass

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

Question

(C) The function is defined as $f(x) = |x^2 - 1|$.We can analyze the behavior of the function by looking at its graph or by examining its critical poi

Vedclass · Accepted Answer

$f$ has a local minima at $x = \pm 1$ and a local maxima at $x = 0$

Vedclass · Answer

(C) The function is defined as $f(x) = |x^2 - 1|$.We can analyze the behavior of the function by looking at its graph or by examining its critical points.$1$. The function $f(x) = |x^2 - 1|$ is non-negative for all $x \in R$.$2$. At $x = 1$ and $x = -1$, $f(x) = |1^2 - 1| = 0$. Since $f(x) \ge 0$ for all $x$, the points $x = 1$ and $x = -1$ are points of local minima where the minimum value is $0$.$3$. At $x = 0$, $f(0) = |0^2 - 1| = |-1| = 1$. For values of $x$ in a small neighborhood around $0$ (e.g.,$x \in (-1, 1)$), $f(x) = |x^2 - 1| = 1 - x^2$. Since $1 - x^2 Thus, $f$ has local minima at $x = \pm 1$ and a local maxima at $x = 0$.

Let $f: R \rightarrow R$ be given by $f(x) = |x^2 - 1|$, then

Similar Questions

On the interval $[1, e]$,the greatest value of $f(x) = x^2 \log x$ is:

Let a rectangle $ABCD$ of sides $2$ and $4$ be inscribed in another rectangle $PQRS$ such that the vertices of the rectangle $ABCD$ lie on the sides of the rectangle $PQRS$. Let $a$ and $b$ be the sides of the rectangle $PQRS$ when its area is maximum. Then $(a+b)^2$ is equal to :

The number of solution$(s)$ satisfying the equation $3x^2 - 2x^3 = \log_2 (x^2 + 1) - \log_2 x$ is

If $x$ is real,then the minimum value of $\frac{x^2-x+1}{x^2+x+1}$ is

The sum of the perimeter of a circle and a square is $k$,where $k$ is a constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module