use the intermediate value theorem to show that f has a zero between a and b.\n\n$f(x)=2x^{3}+6x^{2}-3$; $a…

use the intermediate value theorem to show that f has a zero between a and b.\n\n$f(x)=2x^{3}+6x^{2}-3$; $a = 0$, $b = 1$\n\n$f(a)=\n\n$f(b)=\n\nsince $f(a) ? 0$ and $f(b) ? 0$, we see that $f(c)=0$ for at least one real number c between a and b.

use the intermediate value theorem to show that f has a zero between a and b.\n\n$f(x)=2x^{3}+6x^{2}-3$; $a = 0$, $b = 1$\n\n$f(a)=\n\n$f(b)=\n\nsince $f(a) ? 0$ and $f(b) ? 0$, we see that $f(c)=0$ for at least one real number c between a and b.

Answer

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)