Problem 56 - University Calculus: Early Transcendentals

Question

Does knowing that a function \(g(t)\) is differentiable at \(t=7\) tell you anything about the differentiability of the function \(3 g\) at \(t=7 ?\) Give reasons for your answer.

Step-by-Step Solution

Answer

Yes, if \(g(t)\) is differentiable at \(t=7\), then \(3g(t)\) is also differentiable at \(t=7\) due to scalar multiplication rules.

Step 1: Understanding Differentiability

A function is said to be differentiable at a point if its derivative exists at that point. This implies that the function must be continuous at that point, and it should not have any sharp corners or discontinuities.

Step 2: Analyze Function Modification

The function in question is a scalar multiple of another function, specifically the function is given by \(3g(t)\). When a function is multiplied by a constant, this constant factor impacts the derivative of the function.

Step 3: Apply Derivative Rule for Scalar Multiplication

The derivative of a function \(c \, g(t)\), where \(c\) is a constant, is equal to \(c\) times the derivative of \(g(t)\). Mathematically, \(\frac{d}{dt}[c \, g(t)] = c \, \frac{d}{dt}[g(t)]\). This means that if \(g(t)\) is differentiable at \(t=7\), \(3g(t)\) is also differentiable at \(t=7\).

Step 4: Conclude Differentiability of Modified Function

Since \(g(t)\) is differentiable at \(t=7\), its derivative \(g'(7)\) exists. Consequently, \(3g(t)\) will also have a well-defined derivative at \(t=7\), specifically \(3g'(7)\). Therefore, \(3g(t)\) is differentiable at \(t=7\).

Key Concepts

Derivatives

Understanding derivatives is fundamental in calculus. If you have ever wondered how fast something changes, you're thinking about derivatives. Imagine a car moving down a road. The car's speed is its derivative with respect to time. In the realm of functions, the derivative gives us the slope of the tangent line at any point on the function's graph. You can think of it as a measure of how the function's output changes along its input. Differentiability at a point means that a derivative exists at that point. That implies smoothness at the location—no sudden jumps or sharp turns. If a function is not continuous, it can't have a derivative at that point. Differentiability requires continuity, but the reverse isn't always true: continuous functions might not be differentiable if they bend sharply. The notation for the derivative of a function \( f(x) \) is \( f'(x) \), and it's calculated using limits:\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\] By this formula, we assess how \( f \) changes as \( x \) slightly increases. This makes derivatives a powerful tool for understanding function behavior.

Function Continuity

Function continuity is another cornerstone in understanding calculus and differentiability. A function is continuous if you can draw it without lifting your pencil off the paper. There are no gaps or holes: you move smoothly from one point to the next.For a function \( f(x) \) to be continuous at a point \( x = a \), three things must be true:

• \( f(a) \) is defined; the function gives a real number when you plug in \( a \).
• \( \lim_{x \to a} f(x) \) exists; as you get closer to \( a \), there is a specific value that \( f(x) \) approaches.
• \( \lim_{x \to a} f(x) = f(a) \); the function's value and the limit must match at \( a \).

Continuity is crucial for differentiability; if \( f \) isn't continuous at \( a \), it can't be differentiable there, as the limit processes needed to define \( f'(a) \) require continuity.

Scalar Multiplication in Calculus

Scalar multiplication in calculus refers to multiplying a function by a constant. This operation is essential when examining the impact of constants on the function's slope and rate of change. Consider a function \( f(x) \). When you multiply it by a constant \( c \), it becomes \( c \, f(x) \). Algebraically, this scales the output of the function without changing its input structure.Now, if \( f(x) \) is differentiable, what happens to \( c \, f(x) \) as a result? The derivative of \( c \, f(x) \) is simply \( c \) times the derivative of \( f(x) \). Mathematically, it's expressed as:\[\frac{d}{dx}[c \, f(x)] = c \, \frac{d}{dx}[f(x)]\]This rule shows that the operation maintains differentiability: multiplying by a constant does not disrupt the derivative's existence. It only rescales the rate of change. Suppose \( f(x) \) is differentiable at \( x = a \). Then, \( c \, f(x) \) remains differentiable at \( a \), and its derivative is \( c \, f'(a) \). This insight simplifies complex differentiation problems, as scalar multiples are easy to manage while retaining all essential characteristics of the original function.