How do we use Rolle's Theorem to find whether function #f(x)=4x^2-8x+7# has a point where #f'(x)=0# in the interval #[-1,3]#?

Answer 1

#x=1#, for details please see below.

Rolle's Theorem when applied the function #f(x)# must be continuous for #x# in the given range, here #[-1,3]# and #f(x)# must be differentiable for #x# in #(-1,3)#.
Here we have #a=-1# and #b=3# and as #f(x)=4x^2-8x+7#, we have #f(-1)=19# and #f(3)=19# and hence #f(a)=f(b)#.
Now according to Rolle's Theorem, if in such a function #f(a)=f(b)#, there is one #c#, where #f'(c)=0#
As #f'(x)=8x-8# and as #8x-8=0=>x=1#, we have at #x=1#, #f'(1)=0#
and our #c# is #1#.

graph{(y-4x^2+8x-7)((x+1)^2+(y-19)^2-0.01)((x-3)^2+(y-19)^2-0.01)((x-1)^2+(y-3)^2-0.015)=0 [-4, 4, -5, 23]}

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Answer 2

To use Rolle's Theorem to find whether function f(x)=4x^2-8x+7 has a point where f'(x)=0 in the interval [-1,3], first, we need to verify if the function satisfies the conditions of Rolle's Theorem. Rolle's Theorem states that if a function f(x) is continuous on the closed interval [a,b], differentiable on the open interval (a,b), and f(a) = f(b), then there exists at least one value c in the open interval (a,b) such that f'(c) = 0.

In this case, the function f(x) = 4x^2 - 8x + 7 is a polynomial function, and polynomials are continuous and differentiable for all real numbers. Therefore, f(x) satisfies the conditions of Rolle's Theorem on the interval [-1,3].

Next, we need to find the derivative of the function f(x). The derivative f'(x) represents the rate of change of the function f(x) with respect to x.

f'(x) = d/dx (4x^2 - 8x + 7) = 8x - 8.

Now, we set f'(x) equal to 0 and solve for x to find the critical points:

8x - 8 = 0 => 8x = 8 => x = 1.

So, the critical point occurs at x = 1.

Finally, we need to check if the critical point x = 1 lies within the interval [-1,3]. Since 1 is within the interval [-1,3], we can conclude that there exists at least one point c in the interval [-1,3] such that f'(c) = 0 by Rolle's Theorem. Therefore, the function f(x) = 4x^2 - 8x + 7 has a point where f'(x) = 0 in the interval [-1,3], and that point is x = 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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