Problem 70 - Calculus Single Variable

Question

Suppose that \(f(x), g(x),\) and \(h(x)\) are continuous for \(x \geq a\) and differentiable for \(x>a\). Use the Mean Value Theorem to show that if \(h(a)=0\) and \(0

Step-by-Step Solution

Answer

If \(h\)'s derivative is positive, then \(h(x) > 0\). Similarly, \(g(x) < f(x)\) if \(g'(x) < f'(x)\) and \(g(a) \leq f(a)\).

Step 1: Apply the Mean Value Theorem to h(x)

Since \(h(x)\) is continuous on \([a, b]\) and differentiable on \((a, b)\), the Mean Value Theorem states that there exists a \(c\) in \((a, b)\) such that: \[ h'(c) = \frac{h(b) - h(a)}{b-a} \]Given \(h(a)=0\), we have \[ h'(c) = \frac{h(b)}{b-a} \].

Step 2: Conclude positivity of h(x)

Since \(h'(x) > 0\) for \(a < x\), this implies that \(\frac{h(b)}{b-a} > 0\). Therefore, \(h(b) > 0\) as \(b - a > 0\). Since \(b\) is any point in \(a < x\), we conclude \(h(x) > 0\) for \(a < x\).

Step 3: Relate g(x) and f(x) using the Mean Value Theorem

Apply the Mean Value Theorem to \(g(x)\) and \(f(x)\) under similar conditions: For any \(x > a\), \(g'(c)\) and \(f'(c)\) exist. From the mean value theorem, there exist \(c_g, c_f\) such that: \[ g'(c_g) = \frac{g(x) - g(a)}{x-a}, \quad f'(c_f) = \frac{f(x) - f(a)}{x-a} \].Given \(g(a) \leq f(a)\) and \(g'(x) < f'(x)\), it follows that \( g(x) < f(x) \) for \(x > a\).

Step 4: Final conclusion

Therefore, with the Mean Value Theorem, since \(g(a) \leq f(a)\) and \(g'(x) < f'(x)\), we have \(g(x) < f(x)\) for all \(x > a\). This conclusion is similar to the result obtained for \(h(x)\) using its derivative properties.

Key Concepts

Continuity

Continuity is an essential concept in calculus that ensures a function behaves in a predictable manner without any sudden jumps or breaks. When we say a function like \( h(x) \) is continuous for \( x \geq a \), it means the graph of the function can be drawn without lifting your pencil from the page over this interval. Continuous functions have certain properties that make them favorable for analysis:

• A function is continuous at a point \( a \) if the limit of the function as \( x \) approaches \( a \) is equal to the function's value at \( a \).
• For the Mean Value Theorem to be applicable, a function must be continuous on a closed interval \([a, b]\).
• If a function is continuous on an interval, it will not take sudden changes in value. This allows the Mean Value Theorem to provide meaningful conclusions about the function's behavior.

In our given exercise, the continuity of \( f(x), g(x), \) and \( h(x) \) ensures that these functions do not exhibit sudden breaks over their respective domains, making it possible to use the Mean Value Theorem.

Differentiability

Differentiability is closely related to continuity, but it's a step further. A function is differentiable if it has a derivative at every point in its interval, which implies smoothness and no sharp corners or cusps. For a function to be differentiable on \( (a, b) \), it must also be continuous on that interval, and at every point \( x \), there should be a tangent line:

• If a function is differentiable at a point, it is also continuous at that point. However, not all continuous functions are differentiable.
• Differentiability means you can compute the rate of change or the slope of the function at any point within the interval.
• In our scenario, the differentiability of \( h(x) \) for \( x > a \) confirms that there is a well-defined derivative \( h'(x) \) which we utilize in the Mean Value Theorem.

The assurance of differentiability is what allows the exploration of the behavior of \( h(x) \) using derivatives and the application of the Mean Value Theorem to deduce that \( h(x) > 0 \) for \( a < x \).

Derivative

The derivative is a fundamental tool in calculus that measures how a function's output value changes as its input changes. A derivative provides the slope of the tangent line to the curve at any given point. In mathematical terms, if \( f(x) \) is a function, its derivative \( f'(x) \) is given by:\[ f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}.\]In the exercise, particularly for the function \( h(x) \):

• We're given that \( h'(x) > 0 \) for \( x > a \), which means \( h(x) \) is increasing on this interval. This is crucial information because it tells us that as \( x \) moves away from \( a \), \( h(x) \) gets larger.
• The derivative plays a vital role in the Mean Value Theorem, showing precisely how much the function grows over an interval \([a, b]\).
• In step-by-step logic, knowing \( h'(x) > 0 \) allows us to use the Mean Value Theorem to conclusively state that \( h(x) > 0 \) for all \( x > a \).

Inequalities

Inequalities are mathematical expressions that indicate the relative size or order of two values. In calculus, they often help us bound the behavior of functions or derivatives, as is the case in our exercise. Given conditions involve inequalities:

• \( g(a) \leq f(a) \) at \( x = a \) serves as a starting condition, providing a comparison baseline between \( g \) and \( f \).
• \( g'(x) < f'(x) \) for \( x > a \) implies that after \( x = a \), \( g(x) \) grows at a slower rate than \( f(x) \). This difference in derivatives ensures \( g(x) \) stays under \( f(x) \) as \( x \) increases.
• Working together, these inequalities enable the conclusion that \( g(x) < f(x) \) for \( x > a \), because even if they start equal, the slower rate means \( g(x) \) will fall behind.

Inequalities thus provide the constraints necessary to use derivative comparisons effectively, leading to the meaningful conclusion derived by applying the Mean Value Theorem.