Maximum Value Function Explained
The question asks us to find the specific value of $x$ for which the function $ f(x) = -|x-1|+5 $ achieves its highest possible output. The function is defined for all real numbers ($ \forall x \in \mathbb{R} $).
Absolute Value Component Analysis
To solve this, let's first analyze the components of the function $ f(x) $:
- The term $ |x-1| $ represents the absolute value of the difference between $x$ and $1$.
- A fundamental property of the absolute value function is that it always yields a non-negative result. Thus, for any real number $x$, we have $ |x-1| \geq 0 $.
- The minimum possible value for $ |x-1| $ is $ 0 $.
- This minimum occurs precisely when the expression inside the absolute value bars is zero. We set $ x-1 = 0 $ and solve for $x$: $ x - 1 = 0 $ $ x = 1 $
Function Maximum Calculation
Now, let's examine how the absolute value term affects the overall function $ f(x) = -|x-1|+5 $:
- Since $ |x-1| \geq 0 $, when we multiply it by $ -1 $, the inequality sign flips: $ -|x-1| \leq 0 $. This means the term $ -|x-1| $ will always be zero or negative.
- The largest value that $ -|x-1| $ can possibly take is $ 0 $.
- We already determined that $ -|x-1| $ reaches this maximum value of $ 0 $ when $ x = 1 $.
- Our function $ f(x) $ is constructed by adding $ 5 $ to the term $ -|x-1| $. To find the maximum value of $ f(x) $, we need the term $ -|x-1| $ to be at its maximum.
- The maximum value of $ -|x-1| $ is $ 0 $, and this occurs at $ x = 1 $.
- Substituting $ x = 1 $ into the function gives the maximum value: $ f(1) = -|1-1| + 5 = -|0| + 5 = 0 + 5 = 5 $.
- Therefore, the function $ f(x) $ attains its maximum value of $ 5 $ specifically when $ x = 1 $.
Final Result
The function $ f(x) = -|x-1|+5 $ reaches its maximum value at the $x$-value of $ 1 $.
Important Questions from Algebra (Notes)
- What is the remainder when 2023²⁰²⁴ + 2025²⁰²⁴ is divided by 2024?
In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If she/he attempts all 60 questions and secures 130 marks, the number of questions she/he attempts wrongly, are?
Match List-I with List-II
List-1 List-II (A) If $\begin{bmatrix}\lambda-1 & 0 \\ 0 & \lambda-1 \end{bmatrix} $, then $\lambda$ is (I) 0 (B) If A=$ \begin{bmatrix}1 & 2 \\2 & 4 \end{bmatrix} $, then $\Delta$ is (II) 1 (C) If A = $ \begin{bmatrix}1 & 0 \\0 & \frac{1}{2} \end{bmatrix} $, then $|A^{-1}|$ is (III) -2 (D) If $ \begin{bmatrix}a+1 & 1 \\1 & 2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\1 & 2 \end{bmatrix} $, then a is (IV) 2 Choose the correct answer from the options given below:
- If (x - 1) is a factor of $2x^2 - 5x + k = 0$, then the value of k is:
- If $x = (2+\sqrt{3})^{\frac{1}{3}} + (2+\sqrt{3})^{-\frac{1}{3}}$ and $x^3-3x + k = 0$, then the value of k is:
