Find The Tangent Line To F(x) At X=1

Hey guys! Today, we're diving into the awesome world of calculus to tackle a common problem: finding the equation of a tangent line. Specifically, we've got this function, $f(x) = (2x - 1)^5(x + 1)$, and we need to pinpoint the tangent line at the point where $x=1$. This might sound a bit intimidating with that power of 5, but trust me, with a little bit of differentiation magic, we'll have this solved in no time. Let's break down what a tangent line even is and why it's so crucial in calculus. Think of it as the best linear approximation of a curve at a specific point. It's a line that just touches the curve at that single point and has the same slope as the curve at that exact spot. This concept is fundamental because it allows us to understand the instantaneous rate of change of a function, which has applications in everything from physics (velocity, acceleration) to economics (marginal cost, marginal revenue). So, when we're asked for the equation of a tangent line, we're essentially looking for a line in the form $y = mx + b$, where 'm' is the slope of the line (which will be the derivative of our function evaluated at $x=1$) and 'b' is the y-intercept (which we can find using the point-slope form once we know the slope and the point).

To find the equation of the tangent line, we need two key pieces of information: the slope of the tangent line at $x=1$, and the coordinates of the point on the graph of $f(x)$ at $x=1$. The slope is found by calculating the derivative of the function, $f'(x)$, and then evaluating it at $x=1$. The point is found by simply plugging $x=1$ into the original function, $f(1)$. Let's start with finding the point. When $x=1$, $f(1) = (2(1) - 1)^5(1 + 1) = (2 - 1)^5(2) = (1)^5(2) = 1 imes 2 = 2$. So, the point of tangency is $(1, 2)$. Now, for the slope. This is where we need to use our differentiation rules. The function $f(x)$ is a product of two functions: $u(x) = (2x - 1)^5$ and $v(x) = (x + 1)$. We'll need to use the product rule, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

First, let's find the derivative of $u(x) = (2x - 1)^5$. This requires the chain rule. The outer function is $g(z) = z^5$ and the inner function is $h(x) = 2x - 1$. The derivative of the outer function is $g'(z) = 5z^4$, and the derivative of the inner function is $h'(x) = 2$. So, by the chain rule, $u'(x) = g'(h(x))h'(x) = 5(2x - 1)^4 imes 2 = 10(2x - 1)^4$. Now, let's find the derivative of $v(x) = x + 1$. This is straightforward: $v'(x) = 1$.

Now we can apply the product rule to find $f'(x)$: $f'(x) = u'(x)v(x) + u(x)v'(x)$ $f'(x) = [10(2x - 1)^4](x + 1) + [(2x - 1)^5](1)$

We need the slope at $x=1$, so let's evaluate $f'(1)$: $f'(1) = [10(2(1) - 1)^4](1 + 1) + [(2(1) - 1)^5](1)$ $f'(1) = [10(1)^4](2) + [(1)^5](1)$ $f'(1) = [10(1)](2) + [1](1)$ $f'(1) = 20 + 1$ $f'(1) = 21$

So, the slope of the tangent line at $x=1$ is $m = 21$. We have the point $(1, 2)$ and the slope $m = 21$. Now we can use the point-slope form of a linear equation, which is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is our point and $m$ is our slope.

$y - 2 = 21(x - 1)$

Let's simplify this equation to get it into the $y = mx + b$ form: $y - 2 = 21x - 21$ $y = 21x - 21 + 2$ $y = 21x - 19$

And there you have it! The equation for the line tangent to the graph of $f$ at the point where $x=1$ is $y = 21x - 19$. Let's quickly review the options provided to make sure we match one of them. Option A is $y=21x+2$, Option B is $y=21x-19$, Option C is $y=11x-9$, Option D is $y=10x+2$, and Option E is $y=10x-8$. Our derived equation, $y = 21x - 19$, perfectly matches Option B. It's always a good idea to double-check your work, especially with those chain and product rules, to make sure no little mistakes snuck in. The derivative calculation can be tricky, so breaking it down step-by-step, as we did, is super helpful.

Remember, the tangent line is a powerful tool. It tells us the instantaneous rate of change of the function at a specific point. If this function represented, say, the position of an object, then $f'(1)=21$ would mean the object's velocity at time $x=1$ is 21 units per time unit. The equation of the tangent line itself, $y = 21x - 19$, gives us a linear approximation of the function's behavior near $x=1$. For values of $x$ very close to 1, the value of $y$ on this line will be very close to the value of $f(x)$. This approximation capability is a cornerstone of calculus and is used extensively in numerical methods, optimization, and understanding complex systems. So, mastering these tangent line problems isn't just about passing a test; it's about unlocking a deeper understanding of how functions behave and how we can model the real world using mathematical tools. Keep practicing, and you'll become a calculus pro in no time, guys!

Let's recap the process one more time for good measure. First, we identified the point of tangency by evaluating the function at the given x-value. This gave us our $(x_1, y_1)$ coordinates. Next, we needed the slope, 'm', which is the derivative of the function evaluated at that same x-value. Because our function was a product of two expressions, one of which involved a power, we had to skillfully apply both the product rule and the chain rule. The product rule helped us combine the derivatives of the individual parts, while the chain rule was essential for differentiating the $(2x-1)^5$ term. After calculating $f'(x)$, we plugged in $x=1$ to find the specific slope at our point of interest. Finally, armed with the point and the slope, we used the point-slope form of a linear equation ($y - y_1 = m(x - x_1)$) and rearranged it into the slope-intercept form ($y = mx + b$) to match the format of the answer choices. This systematic approach ensures accuracy and makes even complex calculus problems manageable. So, when you see a function like $f(x)=(2 x-1)^5(x+1)$, don't panic! Just break it down, apply the rules carefully, and you'll be able to find that tangent line like a champ. Math is all about building these foundational skills, and each problem you solve adds another tool to your analytical toolkit. Keep that calculator handy and your thinking cap on!