Question

Graphing \(f\) from \(f^{\prime}\) Sketch the graph of a continuous function \(f\) with \(f(0)=1\) and \(f^{\prime}(x)=\left\\{\begin{array}{ll}{2,} & {x<2} \\\ {-1,} & {x>2}\end{array}\right.\)

Short answer

The function \(f(x)\) increases linearly with a slope of 2 from \(-\infty\) to \(2\), and decreases linearly with a slope of -1 from \(2\) to \(\infty\).

Step-by-step solution

Identify Continuous Segments

This step divides the function into two segments using \(f'(x)\). One is when \(x<2\), \(f'(x)=2\) which is a positively sloped, straight line with slope 2. Another when \(x>2\), \(f'(x)=-1\) which is a negatively sloped, straight line with slope -1.

Sketch \(f'\) based on Slope

Now plot the two segments according to their slope on a plane. The slope expresses the derivative of the function, \(f'(x)\), and tells the direction about whether function is increasing, decreasing or constant.

Integrate to Find \(f(x)\)

Since we know that the derivative of the function gives the slope of the function at a given point, to get the function from its derivative, perform the integral of \(f'(x)\). For the segment with \(f'(x) = 2\), integrate to get \(f(x) = 2x + C_1\). And for \(f'(x)=-1\), then \(f(x) = -x + C_2\). We have to find the constants of integration, \(C_1\) and \(C_2\), using available information.

Use Given Point to Find Constants \(C_1\) and \(C_2\)

Substitute \(f(0)=1\) into the equation \(f(x)=2x+C_1\), then \(C_1 = 1\). From the fact that \(f(x)\) is a continuous function, \(f(2) = f(2)\). That is, \(2(2) + 1 = -2 + C_2\), then \(C_2 = 5\).

Sketch \(f(x)\)

Substitute \(C_1=1\) and \(C_2=5\) back into the equations. For \(x<2, f(x)=2x+1\), and for \(x>2, f(x)= -x+5\). Now, plot the two functions on the same graph with cutoff at \(x=2\).