Question

Let f(x)=5x^3-1 and g(x)=5-4x^2. Evaluate the following. 1. f(x+2)2. (fg)(3)3. (f・f)(2)(Number 4 is a fraction)4. G(x+h)-g(x)/h Show work please!!

Answer 1

Given the functions:

$\begin{gathered} f(x)=5x^3-1 \\ g(x)=5-4x^2 \end{gathered}$

We will find the following:

First, f(x+2)

So, we will substitute with (x+2) into the function f(x)

$f(x+2)=5(x+2)^3-1$

Second, (fg)(3)

so, we will find the product of the functions (f) and (g)

$(fg)(x)=(5x^3-1)\cdot(5-4x^2)$

Then, substitute with x = 3

$(fg)(3)=(5\cdot3^3-1)\cdot(5-4\cdot3^2)=134\cdot(-31)=-4154$

Third, (f・f)(2)

So, we will find the product of the function (f) by (f)

$\begin{gathered} (f\cdot f)(x)=(5x^3-1)\cdot(5x^3-1) \\ (f\cdot f)(x)=(5x^3-1)^2 \end{gathered}$

Then, substitute with x = 2

$(f\cdot f)(2)=(5\cdot2^3-1)^2=(39)^2=1521$

Finally, we will find g(x+h)-g(x)/h

So, we will find g(x+h), then substitute it into the formula.

$g(x+h)=5-4(x+h)^2$

So,

$\frac{g(x+h)-g(x)}{h}=\frac{\lbrack5-4(x+h)^2\rbrack-\lbrack5-4x^2\rbrack}{h}$

Expand the numerator then simplify the answer:

$\begin{gathered} \frac{g(x+h)-g(x)}{h}=\frac{(5-4(x^2+2xh+h^2))-(5-4x^2)}{h} \\ \\ =\frac{5-4x^2-8xh-4h^2-5+4x^2}{h} \\ \\ =\frac{(5-5)+(-4x^2+4x^2)-8xh-4h^2}{h} \\ \\ =\frac{-8xh-4h^2}{h}=\frac{h(-8x-4h)}{h} \\ \\ =-8x-4h \end{gathered}$

So, the answer will be (-8x - 4h)