Let f(x) be a cubic polynomial with f(1) = - 10, f(- 1) = 6, and...

Let, $f\left(x\right) = a{x}^{3} + b{x}^{2} + cx + d$

Diff w. r. t. '$x$'

${f}^{'}\left(x\right) = 3a{x}^{2} + 2bx + c$

Again diff. w. r. t. '$x$'

${f}^{"}\left(x\right) = 6ax + 2b$

${f}^{"}\left(- 1\right) = 0$

$- 6a + 2b = 0$

$b = 3a$

$f ' \left(1\right) = 0$

$3a + 6a + c = 0$

$c = - 9a$

$f\left(1\right) = - 10$

$- 5a + d = - 10 \ldots$\left(i\right)$</p><p>$f ( - 1 ) = 6$</p><p>$11a + d = 6 \ldots \left(ii\right)

From equations \left(i\right) - \left(ii\right)

We get $a = 1, d = - 5, b = 3, c = - 9$

Then, $f\left(x\right)$ = {x}^{3} + 3{x}^{2} - 9x - 5$</p><p>Hence,$f$\left(3\right)$ = 27 + 27 - 27 - 5$</p><p>$f$\left(3\right)$ = 22$</p>$