Question

Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible ) whether they correspond to local maxima or local minima. $$f(x)=e^{x}(x-7)$$

Short answer

Based on the step-by-step solution provided, find the critical points of the function $$f(x) = e^x(x-7)$$ and use the Second Derivative Test to determine if they correspond to local maxima or local minima. Solution: 1. The first derivative of the function is found to be: $$f'(x) = e^x(x-6)$$ 2. The critical point is found to be at $$x=6$$. 3. The second derivative of the function is found to be: $$f''(x) = e^x(x-5)$$ 4. Applying the Second Derivative Test, we find that the critical point $$x=6$$ is a local minimum.

Step-by-step solution

Find the first derivative

To find the first derivative of $$f(x)=e^x(x-7)$$, we will use the product rule, which states: $$(uv)' = u'v + u v'$$ Let \(u=e^x\) and \(v=x-7\). Then, $$u'=e^x$$ and $$v'= 1$$ So, the first derivative is: $$f'(x) = (e^x)(1) + (x-7)(e^x) = e^x(x-6)$$

Find critical points

To find the critical points, we need to solve the equation \(f'(x) = 0\): $$e^x(x-6) = 0$$ Since \(e^x\) is never equal to 0, we can divide both sides by \(e^x\): $$(x-6) = 0$$ Solving this equation, we get $$x=6$$. Thus, we have one critical point at $$x=6$$.

Find the second derivative

To find the second derivative, we will differentiate the first derivative with respect to x. Again, using the product rule, let \(u=e^x\) and \(v=x-6\). Then, $$u'=e^x$$ and $$v'= 1$$ So, the second derivative is: $$f''(x) = (e^x)(1) + (x-6)(e^x) = e^x(x-5)$$

Apply the Second Derivative Test

We will now apply the Second Derivative Test for the critical point found in Step 2. To do this, we need to evaluate the second derivative at the critical point $$x=6$$: $$f''(6) = e^6(6-5) = e^6$$ Since $$e^6$$ is positive, the Second Derivative Test tells us that the critical point $$x=6$$ is a local minimum.

Key concepts

First Derivative

Finding the first derivative of a function is a fundamental step in calculus when dealing with critical points and their nature. The first derivative represents the slope or rate of change of the function. Here, we use the product rule to differentiate the function

• Given: \( f(x) = e^x(x-7) \)
• Apply the product rule: \( (uv)' = u'v + uv' \)
• Identify \( u = e^x \) and \( v = x-7 \)
• Then, \( u' = e^x \) and \( v' = 1 \)

Putting it all together gives us the first derivative of \( f(x) \):\[ f'(x) = (e^x)(1) + (x-7)(e^x) = e^x(x-6) \]Locating critical points involves finding when \( f'(x) = 0 \), which are typically where the slope of the tangent is zero, indicating possible maxima, minima, or points of inflection.

Second Derivative Test

The Second Derivative Test is a tool used to determine whether a critical point of a function is a local maximum, local minimum, or neither. For our function,

• First, we differentiate \( f'(x) = e^x(x-6) \) to find the second derivative.
• Using the product rule again gives: \( f''(x) = (e^x)(1) + (x-6)(e^x) = e^x(x-5) \)

Now, with the second derivative in hand, we evaluate it at the critical point to apply the test: - For the critical point \( x=6 \):Evaluate \( f''(6) = e^6 \).

• If \( f''(x) > 0 \), like here, the function has a local minimum at \( x=6 \).
• If \( f''(x) < 0 \), there would be a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive.

Local Minimum

A local minimum is a point where the function takes a lower value than at any nearby points. To confirm a local minimum using calculus, we utilize the Second Derivative Test. For the function \( f(x) = e^x(x-7) \), we established that \( x=6 \) as the critical point. We applied the Second Derivative Test:

• Calculate \( f''(6) = e^6 \).
• Since \( e^6 \) is positive, it confirms a local minimum exists at \( x=6 \).

Understanding that a local minimum means the graph of the function is concave up at that point (like a cup), which indicates "turning up" back from a decreasing value. This is crucial in optimizing problems where minimizing or maximizing a function is essential. Studying these points helps in applications like economics, physics, and engineering, where optimal values are critical.