91Ó°ÊÓ

The function:

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

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 $-15$.

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

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 localid="1648080572734" $f\left(c\right)=-15$

Note that the Intermediate Value Theorem doesn’t tell us where c is, only that such a c exists somewhere in the interval role="math">[-3,-2]

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 $f\left(c\right)=-15atc≈-2.44$