Answer:
This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3
Step-by-step explanation:
The given function is
When we differentiate this function with respect to x, we get;
We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)
This implies that;
![c-3=\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c-3%3D%5Csqrt%5B3%5D%7B63.15789%7D)
![c=3+\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c%3D3%2B%5Csqrt%5B3%5D%7B63.15789%7D)
If this function satisfies the Mean Value Theorem, then f must be continuous on [1,7] and differentiable on (1,7).
But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.
Answer:
Step-by-step explanation:
Simplify the radical by breaking the radicand up into a product of known factors.
Answer:
x = 90°
Step-by-step explanation:
Using the identity
cos²x = 1 - sin²x
Given
cos²x - sin x + 1 = 0
1 - sin²x - sin x + 1 = 0
- sin²x - sin x + 2 = 0 ← quadratic equation in sin x ( multiply through by - 1 )
sin²x + sin x - 2 = 0 ← in standard form
(sinx + 2)(sinx - 1) = 0 ← in factored form
Equate each factor to zero and solve for x
sinx + 2 = 0 ⇒ sinx = - 2 ← has no solution as - 1 ≤ x ≤ 1
sinx - 1 = 0 ⇒ sinx = 1 ⇒ x = 90°
solution is x = 90°
A. given
b. distributive property (a(b+c)=ab+ac)
c. addition or commutative property (a+b=b+a)
d. subtraction property of equality (if a=b, then a-c=b-c)
e. division property of equality (if a=b, then a/c=b/c)
f. equality property (if a=b, then b=a)
