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)? f(-2.5)≈ (type an integer or decimal rounded to the nearest tenth as needed)

Explanation:

Step1: Use the fundamental theorem of calculus

By the fundamental theorem of calculus, \(f(b)-f(a)=\int_{a}^{b}f^{\prime}(x)dx\). Here \(a = - 3\), \(b=-2.5\), so \(f(-2.5)-f(-3)=\int_{-3}^{-2.5}f^{\prime}(x)dx\).

Step2: Approximate the integral using the graph

The integral \(\int_{-3}^{-2.5}f^{\prime}(x)dx\) can be approximated by the area under the curve \(y = f^{\prime}(x)\) from \(x=-3\) to \(x = - 2.5\).
Looking at the graph, we can approximate this area as a trapezoid. The formula for the area of a trapezoid is \(A=\frac{(y_1 + y_2)}{2}\Delta x\).
When \(x=-3\), \(y=f^{\prime}(-3)=0\) (from the graph). When \(x=-2.5\), \(y = f^{\prime}(-2.5)=1.5\) (approximate value from the graph). And \(\Delta x=-2.5-(-3)=0.5\).
So \(A=\frac{(0 + 1.5)}{2}\times0.5=\frac{1.5\times0.5}{2}=0.375\)

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

Since \(f(-2.5)-f(-3)=\int_{-3}^{-2.5}f^{\prime}(x)dx\) and \(f(-3) = 4\), then \(f(-2.5)=f(-3)+\int_{-3}^{-2.5}f^{\prime}(x)dx\)
Substitute the values: \(f(-2.5)=4 + 0.375=4.4\) (rounded to the nearest tenth)

Answer:

\(4.4\)