Question

refer to the figure below which contains the graph of ( f(x) ), the derivative of the function ( f(x) ). if ( f(-3)=4 ), what is an approximate value of ( f(-2.5) )?

Explanation:

Step1: Use the fundamental theorem of calculus

We know that \(f(x)=f(a)+\int_{a}^{x}f^{\prime}(t)dt\). Here \(a = - 3\) and \(x=-2.5\), so \(f(-2.5)=f(-3)+\int_{-3}^{-2.5}f^{\prime}(t)dt\).

Step2: Approximate the integral using the graph

The integral \(\int_{-3}^{-2.5}f^{\prime}(t)dt\) can be approximated by the area under the curve \(y = f^{\prime}(x)\) from \(x=-3\) to \(x = - 2.5\). Looking at the graph, assume each square has side - length \(1\). The region from \(x=-3\) to \(x=-2.5\) under \(y = f^{\prime}(x)\) is approximately a triangle. The base of the triangle \(b=0.5\) and the height \(h = 1\). The area of a triangle \(A=\frac{1}{2}\times base\times height\). So \(A=\frac{1}{2}\times0.5\times1 = 0.25\)

Step3: Calculate \(f(-2.5)\)

Since \(f(-3) = 4\), then \(f(-2.5)=f(-3)+\int_{-3}^{-2.5}f^{\prime}(t)dt\approx4 + 0.25\)

Answer:

\(4.25\)