The function \( y = f(x) \) is given by the figure.### Graph Description:- **Axes**: The horizontal axis is labeled \( x \) and the vertical axis is labeled \( y \).- **Graph**: The graph is a piecewise linear function composed of several segments: - From \( x = 0 \) to \( x = 1 \), the line is constant at \( y = -1 \). - From \( x = 1 \) to \( x = 2 \), the line increases linearly from \( y = -1 \) to \( y = 1 \). - From \( x = 2 \) to \( x = 4 \), the line is constant at \( y = 1 \). - From \( x = 4 \) to \( x = 5 \), the line decreases linearly from \( y = 1 \) to \( y = -1 \). - From \( x = 5 \) to \( x = 6 \), the line is constant at \( y = -1 \).### Task:Find the minimum and maximum of \( B \) on \([0, 6]\).\[ B(x) = \int_{3}^{x} f(t) \, dt \]

Using the given graph we need to find the maximu and minimum value of:B(x)=∫3x f(t) dtFirst we need to find the critical points:B'(x)=ddx∫3x f(t) dtUsing fundamental theorem of calculas:B'(x)=f(x)Now:B'(x)=0f(x)=0Now in the given graph, f(x) has value zero at: x=1.5 and x=4.5.Therefore the critical points of B(x) are: x=1.5, 4.5Now:B''(x)=0f'(x)=0f'(x) is the slope of function at point x. Now from the given graph, the slope of function at point x=1.5 is positive and slope at point x=4.5 is negative. That is:f'(1.5)>0 and f'(4.5)<0B''(1.5)>0 and B''(4.5)<0Therefore using second derivative test we say that B(x) has minimum value at x=1.5 and maximum value at x=4.5.Minimum value:B(1.5)=∫31.5 f(t) dtB(1.5)=-∫1.53 f(t) dt {Using definite integral property}B(1.5)=-(Area under the curve from x=1.5 to x=3)B(1.5)=-12×0.5×1+1×1B(1.5)=-0.25+1B(1.5)=-1.25Maximum value:B(4.5)=∫34.5 f(t) dtB(4.5)=(Area under the curve from x=3 to x=4.5)B(4.5)=1.25Hence:Minimum value of B is: -1.25Maximum value of B is: 1.25