Q. 61 - 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.

$$\begin{gathered} f\left(x\right)=x^3+2; \\ K=-15 \end{gathered}$$

Step-by-Step Solution

Answer

The approximate value of c of the function at $K=-15$ is $-2.57$$c=-2.44$

Step 1. Given Information.

The function:

$$\begin{gathered} f\left(x\right)=x^3+2; \\ K=-15 \end{gathered}$$

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 $-15$.

$$\begin{gathered} f(-3)={(-3)}^3+2 \\ =-25<-15 \\ f(-2)={(-2)}^3+2 \\ =-6>-15 \end{gathered}$$

Step 3. Apply intermediate value theorem.

Since f is continuous on $[-3,-2]$ and $f(-3)<-15<f(-2)$, by the Intermediate Value Theorem there is some value c ∈ (-3, -2) for which $f\left(c\right)=-15$

Note that the Intermediate Value Theorem doesn’t tell us where c is, only that such a c exists somewhere in the interval style="max-width: none; vertical-align: -4px;" $[-3,-2]$

Step 4. Approximate c.

We can approximate some values of c for which $f\left(c\right)=-15$ by approximating the values
of x for which the graph of $f\left(x\right)=x^3+2$ intersects the line $y=-15$

From this graph we can conclude that $f\left(c\right)=-15 at c\approx -2.44$