Question

use the intermediate value theorem to show that f has a zero between a and b.

$f(x)=2x^{3}+6x^{2}-3$; $a = 0$, $b = 1$

$f(a)=

$f(b)=

since $f(a) ? 0$ and $f(b) ? 0$, we see that $f(c)=0$ for at least one real number c between a and b.

Explanation:

Step1: Calculate f(a)

Substitute \(a = 0\) into \(f(x)=2x^{3}+6x^{2}-3\).
\[f(0)=2\times0^{3}+6\times0^{2}-3=- 3\]

Step2: Calculate f(b)

Substitute \(b = 1\) into \(f(x)=2x^{3}+6x^{2}-3\).
\[f(1)=2\times1^{3}+6\times1^{2}-3=2 + 6-3=5\]

Answer:

\(f(a)=-3\), \(f(b)=5\), since \(f(a)<0\) and \(f(b)>0\)