Let f be a twice-differentiable function such that f(2)=5 and f(5)=2. Let g be the function given by g(x)= f(f(x)).

(a) Explain why there must be a value c for 2 < c < 5 such that f'(c) = -1.

(b) Show that g' (2) = g' (5). Use this result to explain why there must be a value k for 2 < k < 5 such that g"(k)= 0.

(c) Show that if f"(x) = 0 for all x, then the graph of g does not have a point of inflection.

(d) Let h(x) = f(x) - x. Explain why there must be a value r for 2 < r < 5 such that h(r) = 0.

I know you have to use the intermediate value theorem and mean value theorem but don't know how.

(a) Well, if f is twice-differentiable, then we can use the mean value theorem to find such a value c. Since f(2) = 5 and f(5) = 2, we know that there exists a value c in the interval (2, 5) such that:

f'(c) = (f(5) - f(2))/(5 - 2)
= (2 - 5)/(5 - 2)
= -3/3
= -1

So, there must be a value c for 2 < c < 5 such that f'(c) = -1.

(b) Now, let's find the derivatives of g(x) = f(f(x)) using the chain rule. We have:

g'(x) = f'(f(x)) * f'(x)

Since we're interested in finding g'(2) and g'(5), let's evaluate these derivatives at x = 2 and x = 5:

g'(2) = f'(f(2)) * f'(2) = f'(5) * f'(2)
g'(5) = f'(f(5)) * f'(5) = f'(2) * f'(5)

Now, recall that f(2) = 5 and f(5) = 2. So f(f(2)) = f(5) = 2, and f(f(5)) = f(2) = 5. Therefore:

g'(2) = f'(5) * f'(2) = f'(2) * f'(5) = g'(5)

Since g'(2) = g'(5), we can conclude that g''(k) = 0 for some value k in the interval (2, 5). This is because if the first derivatives are equal at two points, then there must exist a point in between where the second derivative is zero, according to the mean value theorem.

(c) If f"(x) = 0 for all x, then the second derivative of g(x) is given by:

g''(x) = [f''(f(x)) * (f'(x))^2] + f'(f(x)) * f''(x)

Since f"(x) = 0 for all x, the second derivative simplifies to:

g''(x) = f'(f(x)) * f''(x)

This means that the second derivative of g(x) is zero for all x. A point of inflection occurs when the second derivative changes sign, but since g''(x) = 0 for all x, the graph of g does not have a point of inflection.

(d) Let's define h(x) = f(x) - x. Notice that h(2) = f(2) - 2 and h(5) = f(5) - 5. Since f(2) = 5 and f(5) = 2, we have:

h(2) = 5 - 2 = 3
h(5) = 2 - 5 = -3

Applying the intermediate value theorem, since h(2) = 3 and h(5) = -3, and h is continuous in the interval [2, 5], there must be a value r in the interval (2, 5) such that h(r) = 0. In other words, there must be a value r for 2 < r < 5 such that h(r) = f(r) - r = 0.

(a) We can use the Intermediate Value Theorem to show that there exists a value c for 2 < c < 5 such that f'(c) = -1. Since f(2) = 5 and f(5) = 2, the function f(x) must pass through every value between 5 and 2 as x varies from 2 to 5. In other words, there exists some c between 2 and 5 such that f(c) = -1 (since -1 is between 5 and 2). By the definition of the derivative, f'(c) is the slope of the tangent line to the graph of f(x) at x = c. Since f(c) = -1, the tangent line to the graph of f(x) at x = c has a slope of -1. Therefore, there must be a value c for 2 < c < 5 such that f'(c) = -1.

(b) To find the derivative of g(x), we can use the Chain Rule. Let h(x) = f(x). Then, g(x) = f(h(x)), so we can write g(x) as g(x) = f(f(x)). Thus, we have:

g'(x) = f'(f(x)) * f'(x).

Now, to show that g'(2) = g'(5), we need to evaluate g'(2) and g'(5). Let's start with g'(2):

g'(2) = f'(f(2)) * f'(2).

Since f(2) = 5, we have:

g'(2) = f'(5) * f'(2).

Similarly, for g'(5), we have:

g'(5) = f'(f(5)) * f'(5).

Since f(5) = 2, we have:

g'(5) = f'(2) * f'(5).

Comparing g'(2) and g'(5), we see that they are equal:

g'(2) = f'(5) * f'(2) = f'(2) * f'(5) = g'(5).

This means that g'(2) and g'(5) have the same value.

Now, if g'(2) = g'(5), by the Mean Value Theorem, there must exist some value k between 2 and 5 such that:

g"(k) = [g'(5) - g'(2)] / (5 - 2) = 0 / 3 = 0.

So, there must be a value k for 2 < k < 5 such that g"(k) = 0.

(c) If f"(x) = 0 for all x, then the second derivative of f(x) is always zero. Let's consider the graph of g(x) = f(f(x)). Since g''(x) represents the second derivative of g(x), we can find it using the Chain Rule:

g''(x) = (f'(f(x)) * f'(x))' = [f''(f(x)) * f'(x) * f'(x) + f'(f(x)) * f''(x)].

Since f''(x) = 0 for all x, we have:

g''(x) = [0 * f'(x) * f'(x) + f'(f(x)) * 0] = 0 + 0 = 0.

This means that g''(x) is always zero, meaning that the graph of g(x) does not have a point of inflection. A point of inflection occurs when the concavity of a function changes, which requires the second derivative to be nonzero.

(d) Let h(x) = f(x) - x. If we want to find a value r for 2 < r < 5 such that h(r) = 0, we need to find a value of x where f(x) = x.

We know that f(2) = 5 and f(5) = 2. Since f(x) is continuous and differentiable on the interval [2, 5], and f(2) > 2 and f(5) < 5, it follows from the Intermediate Value Theorem that there exists a value r for 2 < r < 5 such that f(r) = r.

Therefore, there must be a value r for 2 < r < 5 such that h(r) = f(r) - r = 0.

(a) To prove that there is a value c for 2 < c < 5 such that f'(c) = -1, we can use the Mean Value Theorem. The Mean Value Theorem states that for a function f that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), there exists a value c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).

Since f is twice-differentiable, it is differentiable on the interval (2, 5). Additionally, we know that f(2) = 5 and f(5) = 2. Therefore, we can apply the Mean Value Theorem to the interval (2, 5):

f'(c) = (f(5) - f(2))/(5 - 2)
= (2 - 5)/3
= -1

Therefore, there exists a value c for 2 < c < 5 such that f'(c) = -1.

(b) To show that g'(2) = g'(5), we need to find the derivatives of g(x) and evaluate them at x = 2 and x = 5.

Let's find g'(x) first. By the chain rule, g'(x) = f'(f(x)) * f'(x).

Therefore, g'(2) = f'(f(2)) * f'(2) = f'(5) * f'(2).
g'(5) = f'(f(5)) * f'(5) = f'(2) * f'(5).

From part (a), we know that f'(c) = -1 for some c in (2, 5). Therefore, g'(2) = -1 * f'(2) = -f'(2) and g'(5) = f'(2) * -1 = -f'(2).

Since -f'(2) = -f'(2), we can conclude that g'(2) = g'(5).

Now, to show that there must be a value k for 2 < k < 5 such that g"(k) = 0, we can apply the Mean Value Theorem to g'(x) on the closed interval [2, 5].

The Mean Value Theorem states that if a function g is continuous on [a, b] and differentiable on (a, b), then there exists a value k in (a, b) such that g'(k) = (g(b) - g(a))/(b - a).

Applying this to the interval [2, 5], we have:

g'(k) = (g(5) - g(2))/(5 - 2)
= (g'(5) - g'(2))/3

Using the result from part (b) that g'(5) = g'(2), we have:

g'(k) = (g'(2) - g'(2))/3
= 0/3
= 0

Therefore, there must be a value k for 2 < k < 5 such that g"(k) = 0.

(c) To show that if f"(x) = 0 for all x, then the graph of g does not have a point of inflection, we need to understand the relationship between the second derivative of f and the second derivative of g.

Since g(x) = f(f(x)), we can differentiate g(x) twice using the chain rule:

g'(x) = f'(f(x)) * f'(x)
g"(x) = f"(f(x)) * [f'(x)]^2 + f'(f(x)) * f"(x)

If f"(x) = 0 for all x, then the second term in the expression for g"(x) becomes zero:

g"(x) = f"(f(x)) * [f'(x)]^2 + 0 = [f'(x)]^2 * f"(f(x))

Since f"(x) = 0, we have g"(x) = [f'(x)]^2 * f"(f(x)) = 0 for all x.

Therefore, if f"(x) = 0 for all x, then the graph of g does not have a point of inflection.

(d) Let h(x) = f(x) - x. To show that there must be a value r for 2 < r < 5 such that h(r) = 0, we can use the Intermediate Value Theorem.

The Intermediate Value Theorem states that if a function h is continuous on a closed interval [a, b], and h(a) and h(b) have opposite signs, then there exists at least one value c in (a, b) such that h(c) = 0.

In this case, we have h(2) = f(2) - 2 = 5 - 2 = 3 > 0 and h(5) = f(5) - 5 = 2 - 5 = -3 < 0.

Since h(2) is positive and h(5) is negative, h(a) and h(b) have opposite signs.

Therefore, by the Intermediate Value Theorem, there must be a value r for 2 < r < 5 such that h(r) = 0.

a. use the mean-value theorem

f(2)=5 and f(5)=2
=> (f(5)-f(2))/(5-2)=(2-5)/(5-2)=-1
It is a straight application of the mean-value theorem to state that there exists 2<c<5 such that f'(c)=(f(5)-f(2))/(5-2).

b. Use the chain rule to differentiate g(x) to get
g'(x)=f'(x)*f'(f(x))
So
g'(2)=f'(2)*f'(f(2))=f'(2)*f'(5)
g'(5)=f'(5)*f'(f(5))=f'(5)*f'(2)
Therefore g'(2)=g'(5)

c. If f"(x)=0 ∀x, then
f(x)=mx+b, m,b∈ℝ.
f'(x)=m
g'(x)=m*m(f(x))=m² (m is not dependent on x)
=> g"(x)=0

d. h(x)=f(x)-x
=> h(2)=f(2)-2=5-2=3
and
=> h(5)=f(5)-5=2-5=-3
Since f(x) is continuous and differentiable in (2,5), h(x) is also continuous and differentiable. Since h(r)=0 lies between h(2) and h(5) so the intermediate value theorem applies directly to state that 2<r<5 exists such that h(r)=0.

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