The function of f and g are defined by f(x)=2x+1 and g(x)=5x-1 find fg(x) fg(-3)
Answers
Answer:
Given f (x) = 3x + 2 and g(x) = 4 – 5x, find (f + g)(x), (f – g)(x), (f × g)(x), and (f / g)(x).
To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.
(f + g)(x) = f (x) + g(x)
= [3x + 2] + [4 – 5x]
= 3x + 2 + 4 – 5x
= 3x – 5x + 2 + 4
= –2x + 6
(f – g)(x) = f (x) – g(x)
= [3x + 2] – [4 – 5x]
= 3x + 2 – 4 + 5x
= 3x + 5x + 2 – 4
= 8x – 2
(f × g)(x) = [f (x)][g(x)]
= (3x + 2)(4 – 5x)
= 12x + 8 – 15x2 – 10x
= –15x2 + 2x + 8
\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(
g
f
)(x)=
g(x)
f(x)
= \small{\dfrac{3x+2}{4-5x}}=
4−5x
3x+2
My answer is the neat listing of each of my results, clearly labelled as to which is which.
( f + g ) (x) = –2x + 6
( f – g ) (x) = 8x – 2
( f × g ) (x) = –15x2 + 2x + 8
\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}(
g
f
)(x)=
4−5x
3x+2
f (2) = 2(2) = 4
g(2) = (2) + 4 = 6
h(2) = 5 – (2)3 = 5 – 8 = –3
Now I can evaluate the listed expressions:
(f + g)(2) = f (2) + g(2)
= 4 + 6 = 10
(h – g)(2) = h(2) – g(2)
= –3 – 6 = –9
(f × h)(2) = f (2) × h(2)
= (4)(–3)= –12
(h / g)(2) = h(2) ÷ g(2)
= –3 ÷ 6 = –0.5
Then my answer is:
(f + g)(2) = 10, (h – g)(2) = –9, (f × h)(2) = –12, (h / g)(2) = –0.5
f(x + h)
= 3(x + h)2 – (x + h) + 4
= 3(x2 + 2xh + h2) – x – h + 4
= 3x2 + 6xh + 3h2 – x – h + 4
The expression for the second part of the numerator is just the function itself:
f(x) = 3x2 – x + 4
Now I'll subtract and simplify:
f(x + h) - f(x)f(x+h)−f(x)
= \bigl(3x^2 + 6xh + 3h^2 - x - h + 4\bigr) - \bigl(3x^2 - x + 4\bigr)=(3x
2
+6xh+3h
2
−x−h+4)−(3x
2
−x+4)
= 3x^2 + 6xh + 3h^2 - x - h + 4 - 3x^2 + x - 4=3x
2
+6xh+3h
2
−x−h+4−3x
2
+x−4
= 3x^2 - 3x^2 + 6xh + 3h^2 - x + x - h + 4 - 4=3x
2
−3x
2
+6xh+3h
2
−x+x−h+4−4
= 6xh + 3h^2 - h=6xh+3h
2
−h
All that remains is to divide by the denominator; factoring lets me simplify:
\small{\dfrac{f(x + h) - f(x)}{h}} = \small{\dfrac{6xh + 3h^2 - h}{h}}
h
f(x+h)−f(x)
=
h
6xh+3h
2
−h
= \small{\dfrac{h (6x + 3h - 1)}{h}}=
h
h(6x+3h−1)
= 6x + 3h - 1=6x+3h−1
Now I'm supposed to evaluate at h = 0, so:
6x + 3(0) – 1 = 6x – 1
simplified form: 6x + 3h – 1
value at h = 0: 6x – 1