Question

let (f(x)=(4x^{2}+6x)(6x - 3)). evaluate (f(x)) at the following points (x)-values.
a. (f(12)=)
b. (f(-12)=)

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u(x)v(x)$, then $y'=u'(x)v(x)+u(x)v'(x)$. Let $u(x)=4x^{2}+6x$ and $v(x)=6x - 3$. First, find $u'(x)$ and $v'(x)$.
$u'(x)=\frac{d}{dx}(4x^{2}+6x)=8x + 6$ and $v'(x)=\frac{d}{dx}(6x - 3)=6$.
Then $f'(x)=(8x + 6)(6x - 3)+(4x^{2}+6x)\times6$.

Step2: Expand the expressions

Expand $(8x + 6)(6x - 3)$ and $(4x^{2}+6x)\times6$:
$(8x + 6)(6x - 3)=48x^{2}-24x+36x - 18=48x^{2}+12x - 18$.
$(4x^{2}+6x)\times6 = 24x^{2}+36x$.
So $f'(x)=48x^{2}+12x - 18+24x^{2}+36x=72x^{2}+48x - 18$.

Step3: Evaluate $f'(12)$

Substitute $x = 12$ into $f'(x)$:
$f'(12)=72\times12^{2}+48\times12 - 18$.
$=72\times144+576 - 18$.
$=10368+576 - 18$.
$=10926$.

Step4: Evaluate $f'(-12)$

Substitute $x=-12$ into $f'(x)$:
$f'(-12)=72\times(-12)^{2}+48\times(-12)-18$.
$=72\times144-576 - 18$.
$=10368-576 - 18$.
$=9774$.

Answer:

a. $10926$
b. $9774$