The right option is (c). The Mean Value Theorem does not apply since f(x) is not differentiable on (-1,8). Because the derivative of f is 2/3 x^(-1/3) which is undefined at x=0.So, that would imply that it is not differentiable along the interval.
U=(x+2)
u^2+5u-6=0
C is answer
Answer:
Step-by-step explanation:
We'll take this step by step. The equation is
![8-3\sqrt[5]{x^3}=-7](https://tex.z-dn.net/?f=8-3%5Csqrt%5B5%5D%7Bx%5E3%7D%3D-7)
Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time. First thing is to subtract 8 from both sides:
![-3\sqrt[5]{x^3}=-15](https://tex.z-dn.net/?f=-3%5Csqrt%5B5%5D%7Bx%5E3%7D%3D-15)
The goal is to isolate the term with the x in it, so that means that the -3 has to go. Divide it away on both sides:
![\sqrt[5]{x^3}=5](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E3%7D%3D5)
Let's rewrite that radical into exponential form:

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely. On the right, let's rewrite that back in radical form to solve it easier:
![x=\sqrt[3]{5^5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5E5%7D)
Let's group that radicad into groups of 3's now to make the simplifying easier:
because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:
which is the same as:
![x=5\sqrt[3]{25}](https://tex.z-dn.net/?f=x%3D5%5Csqrt%5B3%5D%7B25%7D)
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Answer:
Given equation have two solutions.
x = 
Right option is (D)
Step-by-step explanation:
We have given,

we can rewrite this equation in the form given as:
6x² + 40 + 31x = 0
or 6x² + 31x + 40 = 0
On solving this quadratic equation, we get :
x = 
That means, given equation have two solutions.
Right option is (D)