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.
Answer:
7
Step-by-step explanation:
x2 + 5x - 7
sub x = 2 in
(2 X 2) + (5 X 2) - 7
4 + 10 - 7
= 7
29, 18, 5, -10, -12, -25
Hope this helped :)
Answer:
I think the answer is 16 cupcakes are not vanilla
Step-by-step explanation:
<u>Answer:</u>
The solution set of given equations -x-y-z = -8 and - 4x + 4y + 5z = 7 and 2x + 2z = 4 is (3, 6, -1)
<u>Solution:</u>
Given, set linear equations are
-x – y – z = -8 ⇒ x + y + z = 8 → (1)
-4x + 4y + 5z = 7 ⇒ 4x – 4y – 5z = -7 → (2)
2x + 2z = 4 ⇒ x + z = 2 → (3)
We have to solve the above given equations using substitution method.
Now take (3), x + z = 2 ⇒ x = 2 – z
So substitute x value in (1)
(1) ⇒ (2 – z) + y + z = 8 ⇒ 2 + y + z – z = 8 ⇒ y + 0 = 8 – 2 ⇒ y = 6.
Now substitute x and y values in (2)
(2) ⇒ 4(2 – z) – 4(6) – 5z = - 7 ⇒ 8 – 4z – 24 – 5z = -7 ⇒ -9z – 16 = -7 ⇒ 9z = 7 – 16 ⇒ 9z = -9 ⇒ z = -1
Now substitute z value in (3)
(3) ⇒ x – 1 = 2 ⇒ x = 2 + 1 ⇒ x = 3
Hence, the solution set of given equations is (3, 6, -1).
