Question

Using the intermediate value​ theorem, determine, if​ possible, whether the function f has at least one real zero between a and b.

Answer 1

`f(x)=x^3+3x^2-6x-16 ~~ \begin{cases} a=-8\\ b=-3 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ f(-8)=(-8)^3+3(-8)^2-6(-8)-16\implies f(-8)=-512+192+48-16 \\\\\\ ~\hspace{19em} f(-8)=-288 \\\\\\ f(-3)=(-3)^3+3(-3)^2-6(-3)-16\implies f(-3)=-27+27+18-16 \\\\\\ ~\hspace{19em} f(-3)=2`

so f(x) changed signs from x = -8 to x = -3, it went from negative to positive, so somewhere in between it had hit the x-axis and thus end up with a solution or root or real zero.