Q. 62 - Calculus

Question

Use the Intermediate Value Theorem to show that for each function f and value K in Exercises 61–66, there must be some c∈R for which f(c) = K. You will have to select an appropriate interval [a, b] to work with. Then find or approximate one such value of c. You may assume that these functions are continuous everywhere.

$$\begin{gathered} f\left(x\right)=-2x^2+4; \\ K=0 \end{gathered}$$

Step-by-Step Solution

Answer

The approximate value of c of the function at $K=0$ is c=1.41

Step 1. Given Information.

The function: 

$$\begin{gathered} f\left(x\right)=-2x^2+4; \\ K=0 \end{gathered}$$

Step 2. Approximate the interval.

By trial and error we can find such values a and b, by testing different values of f(x) until we find one that is less than and one that is greater than $0$

$$\begin{gathered} f(-2)=-2{(-2)}^2+4 \\ =-4<0 \\ f(-1)=-2{(-1)}^2+4 \\ =2>0 \end{gathered}$$

Step 3. Apply intermediate value theorem.

Since f is continuous on $[-2,-1]$ and $f(-2)<0<f(-1)$, by the Intermediate Value Theorem there is some value $c\in [-2,-1]$ for which $f\left(c\right)=0$.

Note that the Intermediate Value Theorem doesn’t tell us where c is, only that such a c exists somewhere in the interval.

Step 4. Approximate c.

We can approximate some values of c for which $f\left(c\right)=0$ by approximating the values
of x for which the graph of $f\left(x\right)=-2x^2+4$ intersects the line $y=0$.

From this graph we can conclude that $f\left(c\right)=0 at c=-1.41$