Let f(x)=4 x^2-4 a x+a^2-2 a+2 and the global minimum value of f(x) for x ∈[0,2] is equal to 3 .

step 1 So, let's solve this question. So in this question, we have F of X is equal to 4x square minus 4

Let f(x)=4 x^2-4 a x+a^2-2 a+2 and the global minimum value...

Let $f(x)=4 x^{2}-4 a x+a^{2}-2 a+2$ and the global minimum value of $f(x)$ for $x \in[0,2]$ is equal to 3 .
The number of values of a for which the global minimum for $\mathrm{x} \in[0,2]$ occurs at the end point of interval $[0,2]$ is
(A) 1
(B) 2
(C) 3
(D) 0

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Quadratic Function and Its Vertex

A quadratic function is a polynomial of degree two that generally forms a parabola when graphed. The vertex of a parabola is the point where the function attains its maximum or minimum value, depending on the orientation of the parabola. Determining the vertex is essential in optimization problems where one is interested in finding the extreme values of the function.

Constrained Optimization on a Closed Interval

When optimizing a function over a closed interval, it is crucial to consider both the interior's critical points and the end points of the domain. In problems where the function's overall minimum or maximum is sought, if the vertex (or other critical points) lies outside the given interval, the optimal value may instead occur at one of the endpoints.

Parameter Influence in Function Behavior

In functions that involve parameters, such as the one given, these parameters affect the position of the vertex and the overall shape of the parabola. Analyzing how the parameter shifts or transforms the function is essential to determining when and where the global minimum or maximum occurs, particularly when the function is subject to constraints such as a specific interval.

Please give Ace some feedback