i am on "rolles and the mean value theorem" and was just wondering, when i am doing rolles, do i really need to find the exact value of x where f'(c) = 0?
for example:
f(x) = (x+4)^2 (x-3) on [-4,3]
i get to:
3x^2+10x-8=7
then i don't know if i then need to find the exact value of x or can i say:
1.2<x<1.3
yes. the mean value theorem lets you pick any interval, but Rolle's theorem says f(a) = f(b). It's just a special case of the MVT where f'(c) = 0. Your interval here meets that condition.
As for your exact value, forget about decimal approximations.
(-5±√70)/3 is exact. f'(1.25) is not zero.
how did you get (-5 +/- sqrt(70))/3
I assumed you wanted the roots of your equation
3x^2+10x-8=7
When applying Rolle's theorem, you actually don't need to find the exact value of x where f'(c) = 0. Instead, you need to show that there exists a value "c" in the interval (-4, 3) where f'(c) = 0.
In your example, you have the equation 3x^2 + 10x - 8 = 7. To show that there exists a value "c" in the interval (-4, 3) where f'(c) = 0, you should follow these steps:
1. Solve the equation 3x^2 + 10x - 8 = 7 by rearranging it into the form: 3x^2 + 10x - 15 = 0.
2. Factorize the equation: (x - 1)(3x + 15) = 0.
3. Set each factor equal to zero: x - 1 = 0 or 3x + 15 = 0.
4. Solve for x in each equation: x = 1 or x = -5.
Now, you have two possible values for "c": c = -5 or c = 1. However, you need to make sure that both of these values lie within the interval (-4, 3).
Checking the interval, you can see that -5 is less than -4, so it is not within the interval (-4, 3). However, 1 is within the interval (-4, 3). Therefore, you can say that there exists a value "c" in the interval (-4, 3) where f'(c) = 0.
In conclusion, you do not need to find the exact value of x where f'(c) = 0 for Rolle's theorem. It is sufficient to show that there exists a value "c" in the interval (-4, 3) where the derivative of the function equals zero. In this case, you have shown that c = 1 satisfies the condition.