Honey: [looking in the cabinet] Hey, where are my cookies?
Kevin: I ate them all.
Honey: What? You ate my cookies?!
Kevin: Yes, but it’s for your own good. You said you’re on a diet, right?
Honey: Yeah, but … that doesn’t mean you can eat my stuff.
Kevin: I just want what’s best for you.
Honey: No, you don’t! Ugh, you’re so full of it!
Winston: Travis, can you do me a favor?
Travis: Sure!
Winston: I need someone to pick up my car from the repair shop.
Travis: No problem.
Winston: Thanks, you’re a lifesaver! Could you do it this afternoon?
Travis: Alright. Just let me know the address, and I’ll be there.
Winston: I’m glad I can count on you. I owe you one!
Lizzie: Hey, I just saw you talking to James. I didn’t know you two were friends!
Hamid: Yeah, we’re pretty close. I’ve known him since I was a kid.
Lizzie: How did you first meet?
Hamid: Actually, we grew up together. He lived next door to us, so he’s kind of a family friend. How do you two know each other?
Lizzie: He’s in my class, but we’re not really close. He’s just a casual acquaintance of mine.
Hamid: You should get to know him better. He’s a great guy. I’m sure you two would really hit it off.
Craig: Cynthia, this is Mister Harris.
Cynthia: Hello Mr. Harris. Pleased to meet you.
Mr. Harris: Please, call me Steve. Mister Harris is my father.
Cynthia: Haha, OK. Thanks, Steve.
Mr. Harris: Have we met before? Your face rings a bell.
Cynthia: I was here for the training session last week.
Mr. Harris: Right, now I remember. I must have seen you there.
——— Mr. Harris excuses himself and leaves ———
Cynthia: Wow, I thought he would be strict, but he’s really kind.
Craig: Yep, Steve is very easygoing. Don’t judge a book by its cover.
Nicole: Hey Wes, long time, no see!
Wesley: It’s good to finally catch up again.
Nicole: How’s your family? Are your folks doing O.K.?
Wesley: They’re fine. My old man is having some trouble with his back, but my mom takes care of him.
Nicole: That’s nice. Say ‘hi’ from me when you see them.
Wesley: I will. You and your better half are still good, right?
Nicole: Yes, we’re still happily married. Thanks for asking.
Richard: Sarah, what a surprise!
Sarah: Hey Richard, fancy meeting you here!
Richard: How have you been?
Sarah: Never been better! I recently got a scholarship at school.
Richard: That’s cool. No wonder you’re so happy.
Sarah: What have you been up to?
Richard: I just graduated from university. So I’ve been looking for a job.
Thomas: Hey Angie, what’s up?
Angie: I’m alright. Nothing special. How’s it going with you?
Thomas: Pretty good. I got a new job recently. So I’ve been busy with that.
Angie: Sounds interesting. Look, I’ve gotta run now, but we should meet up soon.
Thomas: OK, I’ll text you later. Keep in touch.
Angie: You too. Have a good one.
\(f(x)=x^4+x^3-3x^2+1\)의 변곡점에서 접선의 방정식을 구하라.
풀이
\(\displaystyle f'(x)=4x^3+3x^2-6x=x(4x^2+3x-6)\)
\(\displaystyle f'{}'(x)=12x^2+6x-6=6(2x^2+x-1)=6(2x-1)(x+1)\)
\begin{array} {c|c|c|c|c|c} x & \cdots & -1 & \cdots & \frac{1}{2} & \cdots \\ \hline f'{}’ & + & 0 & – & 0 & + \end{array}
\(\displaystyle f(-1)=(-1)^4+(-1)^3-3(-1)^2+1=1-1-3(1)+1=-2\)
\(\displaystyle f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^3-3\left(\frac{1}{2}\right)^2+1\)
\(\displaystyle=\frac{1}{16}+\frac{1}{8}-\frac{3}{4}+1=\frac{1+2-12+16}{16}=\frac{7}{16}\)
변곡점은 \(\displaystyle\left(-1, -2\right), \left(\frac{1}{2}, \frac{7}{16}\right)\)이다.
\(\displaystyle f'(-1)=(-1)(4(-1)^2+3(-1)-6)\)
\(\displaystyle=-(4(1)+-3-6)=-(4-3-6)=-(-5)=5\)
\(\displaystyle f’\left(\frac{1}{2}\right)=\frac{1}{2}\left(4\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)-6\right)\)
\(\displaystyle=\frac{1}{2}\left(4\left(\frac{1}{4}\right)+\frac{3}{2}-6\right)=\frac{1}{2}\left(1+\frac{3}{2}-6\right)\)
\(\displaystyle=\frac{1}{2}\left(\frac{2+3-12}{2}\right)=\frac{1}{2}\left(\frac{-7}{2}\right)=-\frac{7}{4}\)
1. 변곡점 \(\displaystyle\left(-1, -2\right)\)에서 접선의 방정식을 구한다.
\(y-f(-1)=f'(-1)(x-(-1))\)
\(y-(-2)=5(x+1)\)
\(y+2=5x+5\)
\(y=5x+3\)
2. 변곡점 \(\displaystyle\left(\frac{1}{2}, \frac{7}{16}\right)\)에서 접선의 방정식을 구한다.
\(\displaystyle y-f\left(\frac{1}{2}\right)=f’\left(\frac{1}{2}\right)\left(x-\frac{1}{2}\right)\)
\(\displaystyle y-\frac{7}{16}=-\frac{7}{4}\left(x-\frac{1}{2}\right)\)
\(\displaystyle y-\frac{7}{16}=-\frac{7}{4}x+\frac{7}{8}\)
\(\displaystyle y=-\frac{7}{4}x+\frac{7}{8}+\frac{7}{16}\)
\(\displaystyle y=-\frac{7}{4}x+\frac{14+7}{16}\)
\(\displaystyle y=-\frac{7}{4}x+\frac{21}{16}\)
이계도함수 판정법을 이용하여 다음 함수의 극점을 판별하여라.
$$f(x)=x\sqrt{8-x^2}$$
풀이
\(\displaystyle f(x)=x(8-x^2)^\frac{1}{2}\)
\(\displaystyle f'(x)=x'(8-x^2)^\frac{1}{2}+x\left((8-x^2)^\frac{1}{2}\right)’\)
\(\displaystyle=(8-x^2)^\frac{1}{2}+x\left(\frac{1}{2}(8-x^2)^{-\frac{1}{2}}(-2x)\right)\)
\(\displaystyle=(8-x^2)^\frac{1}{2}-x^2(8-x^2)^{-\frac{1}{2}}\)
\(\displaystyle=\sqrt{8-x^2}-\frac{x^2}{\sqrt{8-x^2}}\)
\(\displaystyle=\frac{8-x^2-x^2}{\sqrt{8-x^2}}=\frac{8-2x^2}{\sqrt{8-x^2}}\)
\(\displaystyle=\frac{-2(x^2-4)}{\sqrt{8-x^2}}=\frac{-2(x+2)(x-2)}{\sqrt{8-x^2}}\)
\(\displaystyle f'{}'(x)=\frac{(8-2x^2)’\sqrt{8-x^2}-(8-2x^2)(\sqrt{8-x^2})’}{8-x^2}\)
\(\displaystyle=\frac{-4x\sqrt{8-x^2}-(8-2x^2)((8-x^2)^\frac{1}{2})’}{8-x^2}\)
\(\displaystyle=\frac{-4x\sqrt{8-x^2}-(8-2x^2)(\frac{1}{2}(8-x^2)^{-\frac{1}{2}}(-2x))}{8-x^2}\)
\(\displaystyle=\frac{-4x\sqrt{8-x^2}+x(8-2x^2)((8-x^2)^{-\frac{1}{2}})}{8-x^2}\)
\(\displaystyle=\frac{-4x\sqrt{8-x^2}+\frac{x(8-2x^2)}{\sqrt{8-x^2}}}{8-x^2}\)
\(\displaystyle=\frac{\frac{-4x(8-x^2)+x(8-2x^2)}{\sqrt{8-x^2}}}{8-x^2}\)
\(\displaystyle=\frac{-4x(8-x^2)+x(8-2x^2)}{\sqrt{8-x^2}(8-x^2)}\)
\(\displaystyle=\frac{4x^3-32x-2x^3+8x}{\sqrt{8-x^2}(8-x^2)}\)
\(\displaystyle=\frac{2x^3-24x}{\sqrt{8-x^2}(8-x^2)}\)
\(\displaystyle f'(x)=0\)인 \(x\)의 값은 \(-2\), \(2\)이다.
\(\displaystyle f'{}'(-2)=\frac{2(-2)^3-24(-2)}{\sqrt{8-(-2)^2}(8-(-2)^2)}\)
\(\displaystyle=\frac{2(-8)+48}{\sqrt{8-4}(8-4)}=\frac{-16+48}{\sqrt{4}(4)}\)
\(\displaystyle=\frac{32}{2(4)}=\frac{32}{8}=4>0\)
\(\displaystyle f'{}'(2)=\frac{2(2)^3-24(2)}{\sqrt{8-(2)^2}(8-(2)^2)}\)
\(\displaystyle=\frac{2(8)-48}{\sqrt{8-4}(8-4)}\)
\(\displaystyle=\frac{16-48}{\sqrt{4}(4)}=\frac{-32}{2(4)}\)
\(\displaystyle=\frac{-32}{8}=-4<0\)
\(f(-2)=(-2)\sqrt{8-(-2)^2}=-2\sqrt{8-4}=-2\sqrt{4}=-2(2)=-4\)
\(f(2)=(2)\sqrt{8-(2)^2}=2\sqrt{8-4}=2\sqrt{4}=2(2)=4\)
따라서 \(f(x)\)는 \(x=-2\)일 때 극소값 \(-4\), \(x=2\)일 때 극대값 \(4\)를 갖는다.
이계도함수 판정법을 이용하여 다음 함수의 극점을 판별하여라.
$$f(x)=8x^2-x^4$$
풀이
\(f'(x)=-4x^3+16x=-4x(x^2-4)=-4x(x+2)(x-2)\)
\(f'{}'(x)=-12x^2+16=-4(3x^2-4)\)
\(f'(x)=0\)인 \(x\)의 값은 \(-2\), \(0\), \(2\)이다.
\(f'{}'(-2)=-4(3(-2)^2-4)=-4(3(4)-4)=-4(12-4)\)
\(=-4(8)=-32<0\)
\(f'{}'(0)=-4(3(0)^2-4)=-4(-4)=16>0\)
\(f'{}'(2)=-4(3(2)^2-4)=-4(3(4)-4)=-4(12-4)\)
\(=-4(8)=-32<0\)
\(f(x)=8x^2-x^4=x^2(8-x^2)\)
\(f(-2)=(-2)^2(8-(-2)^2)=4(8-4)=4(4)=16\)
\(f(0)=(0)^2(8-(0)^2)=0(8-0)=0\)
\(f(2)=(2)^2(8-(2)^2)=4(8-4)=4(4)=16\)
따라서 \(f(x)\)는 \(x=-2\)일 때 극대값 \(16\), \(x=0\)일 때 극소값 \(0\), \(x=2\)일 때 극대값 \(16\)을 갖는다.