Hey, everyone. In this problem, we're asked to answer the questions below given the function \( f(x) = 2x^3 - 3x + 5 \). So let's dig into these questions. Our first question, a, asks us to find \( f'(x) \), the derivative of our function \( f(x) \). I'm gonna go ahead and do my work for this down here.
So here, starting with part (a), I want to find the derivative \( f'(x) \). Now, looking at my original function, I know that I can go ahead and use the power rule on this polynomial. Now using the power rule here, I want to pull that 3 out to the front, decrease that exponent by 1. That gives me here \( 6x^{3-1} \), which is just 2 because I've multiplied that 2 times that 3 that I pulled out to the front and decreased that power by 1. Now looking at my next term here, I have \(-3x\).
Now whenever I have a constant multiplying \( x \), I know that my derivative is just going to be that constant by itself because the derivative of \( x \) is just 1. So this becomes a \(-3\). Then my last term is just a constant. The derivative of any constant is 0. So I already have my final answer here.
The derivative \( f'(x) \) is \( 6x^2 - 3 \). So we're done with part (a). Let's go ahead and move on to part (b). Now, in part (b), we're asked to find the equation of the tangent line at \( x = 1 \). So what information do we need in order to be able to do this?
Well, I know that I can set up my point-slope formula, \( y - y_1 = m(x - x_1) \). So based on this point-slope formula, what is all of the information that we need in order to find the equation of this tangent line? Well, I need \( y_1 \), I need \( m \), and I need \( x_1 \). So what do we need to find here? Well, I already know that \( x_1 \) is going to be one because that's already given to me here.
This is my \( x_1 \) value. Now so I need to find \( m \), which is my slope, and I need to find \( y_1 \), which is just the value of my original function with \( x_1 \) plugged in. So let's go ahead and start there with \( y_1 \). Now, in order to find \( y_1 \), we just need to plug 1 into our original function. So this is \( f(1) \).
Now plugging that into our original function, which is \( 2x^3 - 3x + 5 \), I have \( 2 \times 1^3 - 3 \times 1 + 5 \). Now this ends up giving me \( 2 - 3 + 5 \), which if I subtract and add there, this is going to end up giving me a positive 4. So this is my \( y_1 \) value that is ready to plug into that point-slope formula. But we need to find one more thing here. That's our \( m \), our slope value.
Remember that the derivative of a function is the slope of its tangent line. So, in order to find the slope of our tangent line specifically at \( x = 1 \), we just need to plug 1 into our derivative. Now we already found our derivative in part (a), so in order to find \( m \) here I'm just going to find \( f'(1) \), plugging in that \( x \) value that we're given. So \( f'(1) \), having found my derivative in part (a), this is \( 6 \times 1^2 - 3 \). Now \( 6 \times 1^2 \) is just 6, so this is \( 6 - 3 \), which is just 3, and this is my \( m \) value.
Now from here, we can plug all of this into our point-slope formula. So, scrolling down a little bit here since I need a little bit of space, plugging all these values in, I have \( y - y_1 \), which here is 4. So plugging that 4 in, \( y - 4 = m(x - x_1) \), my slope, which we found was 3. So \( 3(x - x_1) \), which again was already given to us. That was our value of 1 given in our problem statement.
Now we can simplify this because we don't typically present equations of lines in this way. So I'm gonna go ahead and distribute this 3 into these parentheses and also move this 4 over to this side. So this becomes \( y = 3x - 3 \), distributing that 3, plus 4. Now \(-3 + 4\) is just positive 1. So my final answer here for the equation of my tangent line at \( x = 1 \) is \( y = 3x + 1 \).
Now we've completed part (b). We have one final question here, part (c), which asks us to find the values of \( x \) for which the tangent line is horizontal. Now how exactly can we figure this out? What does that even mean, the tangent line is horizontal? Well, when we think of a horizontal line, remember that the slope of a horizontal line is 0.
Now, we know that the slope of our tangent line is our derivative. Since we found our derivative, we can take that derivative \( f'(x) \) and set it equal to 0 in order to find all of those \( x \) values for which our tangent line is horizontal. So let's go ahead and plug this in. I have \( 6x^2 - 3 \). That's my derivative here that I am setting equal to 0.
Now if I actually solve here, I can add 3 to both sides. Canceling it out here, I have \( 6x^2 = 3 \). Dividing by 6 on both sides gives me \( x^2 = \frac{3}{6} \), which is just \(\frac{1}{2}\). Then my final step here is to take the square root. So my final answer here is \( x = \pm \sqrt{\frac{1}{2}} \).
Remember, since we're taking the square root, we want to account for all of those values, positive or negative. Now the square root of \(\frac{1}{2}\) you may also see this written as \(\frac{\sqrt{2}}{2}\), so this is also an acceptable answer, \(\pm \frac{\sqrt{2}}{2}\) or \(\pm \sqrt{\frac{1}{2}}\). Now we're finally all done here. I know this was a rather tedious problem, but we want to get as much practice as possible here. Thanks for watching, and let me know if you have any questions.
