What is the value of b?
Consider the following for the next two (02) items that follow: Let f(x)=\begin{cases}\frac{x-3}{|x-3|}+a,&x \lt 3\\ a-b,&x=3\\ \frac{x-3}{|x-3|}+b,&x \gt 3\end{cases}. The function f(x) is continuous at x = 3.
- A. -1
- B. 1 ✓
- C. 2
- D. 3
Correct Answer: B. 1
Explanation
From the continuity condition a-1 = b+1 = a-b established in the previous question, solving a-1 = a-b directly yields b=1.
Related questions on Calculus
- Let z=[y] and y=[x]-x, where [.] is the greatest integer function. If x is <strong>NOT</strong> an integer but positive, then what i...
- If f(x)=4x+1 and g(x)=kx+2 such that fog(x)=gof(x), then what is the value of k?
- What is the <strong>MINIMUM</strong> value of the function f(x)=\log_{10}(x^{2}+2x+11)?
- What is \int(x^{x})^{2}(1+\ln x)\,dx equal to ?
- What is \int e^{x}\{1+\ln x+x\ln x\}\,dx equal to?