Answer:
41. f⁻¹(x) = -9x + 4
43. m⁻¹(x) = ∛(x-2)/4
Step-by-step explanation:
41. y = (4-x)/9
swap x and y: x = (4-y)/9
solve y: 9x = 4-y
y = -9x + 4
45. y = 4x³+2
x = 4y³+2
4y³ = x-2
y³ = (x-2)/4
y = ∛(x-2)/4
Let you do 42 and 46 by yourself
Answer:
(x+1)^2 + (y-3)^2 = 16
Step-by-step explanation:
The equation of a circle is given by
(x-h)^2 + (y-k)^2 = r^2
Where (h,k) is the center and r is the radius
(x--1)^2 + (y-3)^2 = 4^2
(x+1)^2 + (y-3)^2 = 4^2
(x+1)^2 + (y-3)^2 = 16
Solve for <em>x</em> when √(<em>x</em> ² - 4) = 1 :
√(<em>x</em> ² - 4) = 1
<em>x</em> ² - 4 = 1
<em>x</em> ² = 5
<em>x</em> = ±√5
We're looking at <em>x </em>≤ 0, so we take the negative square root, <em>x</em> = -√5.
This means <em>f</em> (-√5) = 1, or in terms of the inverse of <em>f</em>, we have <em>f</em> ⁻¹(1) = -√5.
Now apply the inverse function theorem:
If <em>f(a)</em> = <em>b</em>, then (<em>f</em> ⁻¹)'(<em>b</em>) = 1 / <em>f '(a)</em>.
We have
<em>f(x)</em> = √(<em>x</em> ² - 4) → <em>f '(x)</em> = <em>x</em> / √(<em>x</em> ² - 4)
So if <em>a</em> = -√5 and <em>b</em> = 1, we get
(<em>f</em> ⁻¹)'(1) = 1 / <em>f '</em> (-√5)
(<em>f</em> ⁻¹)'(1) = √((-√5)² - 4) / (-√5) = -1/√5
The sign must be negative; see the attached plot, and take note of the negatively-sloped tangent line to the inverse of <em>f</em> at <em>x</em> = 1.