Q. 12 - Calculus

Question

Suppose f is a function whose derivative $f\text{'}$ is given by the graph shown next. In Exercises 9–12, use the given value of f , an area approximation, and the Fundamental Theorem to approximate the requested value

Given that $f(-1)=2$, approximate $f\left(1\right)$.

Step-by-Step Solution

Answer

The required approximation $f\left(1\right)=5.17$

Step 1. Given information

Given  that $f(-1)=2$ and graph of $f\text{'}$

Step 2. Finding the value of f ( 1 )

Use the method. If f is derivative of the function $f\text{'}$ then they by fundamental theorem $f\left(x\right)=\int f\text{'}\left(x\right) dx+C$

The graph (parabola) of the function shows that $f\text{'}$ is quadratic function whose roots are 2.5 and -0.5

Therefore,$$\begin{gathered} f\text{'}\left(x\right)=(x-2.5)(x-0.5)=x^2-3x+1.25 \\ Now,f\left(x\right)=f\text{'}\left(x\right) dx+C, where C is a cons\tan t \end{gathered}$$

$$\begin{gathered} \Rightarrow f\left(x\right)=\int \left(x^2-3x+1.25\right)dx+C \\ \Rightarrow f\left(x\right)=\frac{x^3}{3}-3\frac{x^2}{2}+1.25x+C............\left(1\right) \end{gathered}$$

we have $f(-1)=2...........\left(2\right)$

From results (1) and (2)

$$\begin{gathered} f(-1)=\frac{{\left(-1\right)}^3}{3}-3\frac{{(-1)}^2}{2}+1.25(-1)+C \\ \Rightarrow f(-1)=-\frac{1}{3}-\frac{3}{2}-\frac{5}{4}+C \\ \Rightarrow 2=-\frac{1}{3}-\frac{3}{2}-\frac{5}{4}+C \\ C=2+\frac{1}{3}+\frac{3}{2}+\frac{5}{4} \\ =\frac{24+4+18+15}{12} \\ =\frac{61}{12} \end{gathered}$$

Substitute value of C in result (1)

$f\left(x\right)=\frac{x^3}{3}-3\frac{x^2}{2}+\frac{5}{4}x+\frac{61}{12}.......................\left(3\right)$

$$\begin{gathered} f\left(1\right)=\frac{{\left(1\right)}^3}{3}-3\frac{{\left(1\right)}^2}{2}+\frac{5}{4}\left(1\right)+\frac{61}{12} \\ =\frac{1}{3}-\frac{3}{2}+\frac{5}{4}+\frac{61}{12} \\ \Rightarrow f\left(1\right)=\frac{4-18+15+61}{12} \\ =\frac{62}{12} \\ =5.17 \end{gathered}$$

Thus,$f\left(1\right)=5.17$