Given that f(x) = (x+1)^2(3x-4)?
Expand the exponent first. f(x) = (x² + 2x + 1)(3x - 4) f(x) = 3x³ + 6x² + 3x - 4x² - 8x - 4 f(x) = 3x³ + 2x² - 5x - 4
Given f(x)=x^2-3X+4, find f(x+2)-f(2)?
f(x) = x² - 3x + 4 f(x+2) - f(2) = (x+2)² - 3(x+2) + 4 - (2² - 3*2 + 4) = x² + 4x + 4 - 3x - 6 + 4 - 4 + 6 - 4 = x² + x(4-3) + (4-6+4-4+6-4) = x² + x = x(x+1)
Given F(x) = 3x 2 - 4 and F(a) = 8, find a.?
F(x) = 3x ^2 - 4 Plug in a. F(a) = 3a^2 - 4 ___________________________________ F(a)=8, so 3a^2 - 4 =8 or, 3a^2 - 12 =0 a^2 - 4 =0 a^2 =4 |a| = 2 a = ±2
Find f^ -1(x) if f(x) = 3x + 4.?
reverse x and y x=3y+4 y=(x-4)/3
Given the polynomial F(x) = 3x^4 - 3x + 3, find F(2).?
F(x) = 3x^4 - 3x + 3 F(2) = 3(2)^4 - 3(2) + 3 = 3(16) - 6 + 3 = 48 - 3 = 45
If f(x) =3x^2 and g(x) =x-1, how do you find f(g(x)) and g(f(x))?
These problems use the 'nested' function rule. If f(x) = 3x^2, and g(x) = x - 1:f(g(x)) means that the function g(x) itself will be used as an 'x' value in f(x)likewise, in g(f(x)) means that the function f(x) itself will be used as an 'x' value in g(x)Hence, f(g(x)) = f(x-1) = 3(x-1)^2.And g(f(x)) = g(3x^2) = (3x^2) - 1
How can I find [math]f(2)[/math] and [math]f(a+h)[/math] when [math]f(x) =3x^2+2x+4[/math]?
The important thing to remember about questions like this is that when you see a formula like [math]f(x)=[/math](formula involving [math]x[/math]), it means that whatever you find between the parentheses on the left side is used to replace every occurrence of x on the right side. Furthermore, it’s good practice to keep the parentheses when you make those replacements.So, with [math]f(x)=3x^2+2x+4[/math]:[math]\begin{align*}f(2)&=3(2)^2+2(2)+4=12+4+4=20\\[5pt] f(a+h)&=3(a+h)^2+2(a+h)+4\\&=3a^2+6ah+3h^2+2a+2h+4\\[5pt] &\text{and just for fun}\dots\\ f(1+\sqrt3)&=3(1+\sqrt3)^2+2(1+\sqrt3)+4\\&=3+6\sqrt3+9+2+2\sqrt3+4\\&=18+8\sqrt3\end{align*}\tag*{}[/math]Important note: Do not fall into the trap—a common mistake for students at the precalculus/calculus level—of thinking that, e.g., [math]f(x+1)=f(x)+1[/math]. The correct answer would be:[math]\begin{align*}\qquad f(x+1)&=3(x+1)^2+2(x+1)+4\\&=3x^2+6x+3+2x+2+4\\&=3x^2+8x+9\end{align*}\tag*{}[/math]… which happens to be [math]f(x)+6x+5[/math], but I don’t know of any general shortcuts to get there for arbitrary functions. (One can find shortcuts for specific types of functions, as Xavier Dectot did for quadratic functions like this one.)How can I find f(2) and f(a+h) when f(x) =3x^2+2x+4?