Question

Let $f(x)=10-|x-3|,\; x\in\mathbb{R}$. The maximum of $f(x)$ occurs at:

• A. $x=0$
• B. $x=3$
• C. $x=-3$
• D. $x=10$
• E. $x=1$

Hint:

Functions of the form $f(x) = K - |x-a|$ always reach their maximum value $K$ at $x=a$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to find the value of x for which the function \( f(x) \) attains its maximum value. The function involves an absolute value term, which is always non-negative.
Step 2: Key Formula or Approach:
The function is given by \( f(x) = 10 - |x-3| \). To maximize \( f(x) \), we need to subtract the smallest possible value from 10. The term being subtracted is \( |x-3| \). The value of an absolute value expression is always greater than or equal to zero.
\[ |x-3| \geq 0 \] Step 3: Detailed Explanation:
To make \( f(x) \) as large as possible, we need to make the term \( |x-3| \) as small as possible.
The minimum value of \( |x-3| \) is 0.
This minimum occurs when the expression inside the absolute value is zero:
\[ x - 3 = 0 \] \[ x = 3 \] At \( x=3 \), the value of the function is:
\[ f(3) = 10 - |3-3| = 10 - |0| = 10 \] For any other value of x, \( |x-3| \) will be a positive number, which means \( f(x) \) will be less than 10. For example, if \( x=4 \), \( f(4) = 10 - |4-3| = 10 - 1 = 9 \).
Thus, the maximum value of \( f(x) \) is 10, and this occurs at \( x=3 \).
Step 4: Final Answer:
The maximum of f(x) occurs at x=3.