Cubic Function Extrema: Min, Max, And Values

Hey guys! Today, we're diving deep into the fascinating world of calculus to tackle a common problem: finding the local minimum and local maximum of a cubic function. We've got a specific function on our hands, $f(x) = 2x^3 - 27x^2 + 48x + 5$. This bad boy is guaranteed to have one local minimum and one local maximum, and our mission is to pinpoint exactly where these occur and what the function's value is at those points. So, grab your calculators and your thinking caps, because this is going to be a fun ride!

Unlocking Local Extrema with Derivatives

Alright, so how do we actually find these elusive local minimums and maximums? The secret sauce lies in the first derivative of the function. Remember, the derivative of a function, $f'(x)$, tells us the slope of the tangent line at any given point. At a local minimum or a local maximum, the tangent line is perfectly horizontal, meaning its slope is zero. Therefore, to find our critical points (where these extrema might occur), we need to find the values of $x$ where $f'(x) = 0$. Let's get our hands dirty with our specific function, $f(x) = 2x^3 - 27x^2 + 48x + 5$. First, we need to compute its derivative. Using the power rule for differentiation, which states that the derivative of $ax^n$ is $n imes ax^{n-1}$, we can differentiate each term:

• The derivative of $2x^3$ is $3 imes 2x^{3-1} = 6x^2$.
• The derivative of $-27x^2$ is $2 imes -27x^{2-1} = -54x$.
• The derivative of $48x$ is $1 imes 48x^{1-1} = 48x^0 = 48$.
• The derivative of a constant, 5, is 0.

So, our first derivative, $f'(x)$, is $6x^2 - 54x + 48$. Now, we set this derivative equal to zero to find our critical points: $6x^2 - 54x + 48 = 0$. This is a quadratic equation, and we can solve it by factoring or using the quadratic formula. Let's try factoring first. We can simplify the equation by dividing the entire equation by 6: $x^2 - 9x + 8 = 0$. Now, we need to find two numbers that multiply to 8 and add up to -9. Those numbers are -1 and -8. So, we can factor the quadratic as $(x - 1)(x - 8) = 0$. Setting each factor to zero gives us our critical points: $x - 1 = 0 igh tarrow x = 1$ and $x - 8 = 0 igh tarrow x = 8$. These are the $x$-values where our function might have a local minimum or maximum. But we're not done yet; we need to determine which is which!

The Second Derivative Test: Confirming Minima and Maxima

So, we've found our critical points at $x = 1$ and $x = 8$. But are they local minimums or maximums? This is where the second derivative test comes in handy, guys. The second derivative, $f''(x)$, tells us about the concavity of the original function. If the second derivative is positive at a critical point, the function is concave up (like a smiley face 😊), indicating a local minimum. If the second derivative is negative, the function is concave down (like a frowny face ☹️), indicating a local maximum. If the second derivative is zero, the test is inconclusive, and we'd have to use the first derivative test, but usually, for these types of problems, the second derivative test does the trick. Let's find the second derivative of our function. We differentiate our first derivative, $f'(x) = 6x^2 - 54x + 48$:

• The derivative of $6x^2$ is $2 imes 6x^{2-1} = 12x$.
• The derivative of $-54x$ is $1 imes -54x^{1-1} = -54x^0 = -54$.
• The derivative of 48 is 0.

So, our second derivative is $f''(x) = 12x - 54$. Now, let's plug in our critical points:

• At $x = 1$: $f''(1) = 12(1) - 54 = 12 - 54 = -42$. Since $-42$ is negative, the function is concave down at $x = 1$. This means we have a local maximum at $x = 1$.
• At $x = 8$: $f''(8) = 12(8) - 54 = 96 - 54 = 42$. Since $42$ is positive, the function is concave up at $x = 8$. This means we have a local minimum at $x = 8$.

Boom! We've successfully identified the locations of our local minimum and maximum. The local minimum is at $x=8$, and the local maximum is at $x=1$. But we're not quite done, we still need to find the actual function values at these points!

Calculating Function Values at Extrema

We've figured out where the local minimum and maximum occur ($x=8$ and $x=1$, respectively). The final step is to determine the function value at these specific $x$-coordinates. This is where we plug our $x$-values back into the original function, $f(x) = 2x^3 - 27x^2 + 48x + 5$. Let's do this step-by-step:

• For the local maximum at $x = 1$: $f(1) = 2(1)^3 - 27(1)^2 + 48(1) + 5$ $f(1) = 2(1) - 27(1) + 48 + 5$ $f(1) = 2 - 27 + 48 + 5$ $f(1) = -25 + 48 + 5$ $f(1) = 23 + 5$ $f(1) = 28$ So, the local maximum value of the function is 28, occurring at $x=1$.

• For the local minimum at $x = 8$: $f(8) = 2(8)^3 - 27(8)^2 + 48(8) + 5$ $f(8) = 2(512) - 27(64) + 384 + 5$ $f(8) = 1024 - 1728 + 384 + 5$ $f(8) = -704 + 384 + 5$ $f(8) = -320 + 5$ $f(8) = -315$ Therefore, the local minimum value of the function is -315, occurring at $x=8$.

The Grand Finale: Summarizing Our Findings

We've done it, guys! We've successfully analyzed the function $f(x) = 2x^3 - 27x^2 + 48x + 5$ and found its local extrema. By using the power of derivatives, specifically the first and second derivative tests, we've pinpointed the exact locations and values of the local minimum and maximum.

To recap:

• We found the critical points by setting the first derivative, $f'(x) = 6x^2 - 54x + 48$, equal to zero, which gave us $x=1$ and $x=8$.
• We used the second derivative, $f''(x) = 12x - 54$, to classify these critical points. $f''(1) = -42$ (negative) indicated a local maximum, and $f''(8) = 42$ (positive) indicated a local minimum.
• Finally, we plugged these $x$-values back into the original function to find the corresponding function values. The local maximum is at $x=1$ with a function value of $f(1)=28$. The local minimum is at $x=8$ with a function value of $f(8)=-315$.

So, to fill in those blanks, the function has a local minimum at $x=$ 8 with function value -315 and a local maximum at $x=$ 1 with function value 28. Pretty neat, huh? Calculus is a powerful tool for understanding the behavior of functions, and finding these turning points is a fundamental skill. Keep practicing, and you'll be a calculus whiz in no time!