Hi, there!
When I first went on a business trip to England, I was “flabbergasted” about
how courteously the taxi driver served me at the Heathrow Airport terminal.
“Where would you like to go, sir?” said a driver.
“Sir!?” He greeted a young Japanese stuck-up brat in my early twenties.
“To Hotel so-and-so, please,” I answered politely.
Then, I asked him to put my luggage in the trunk, but my tone of voice might
have been unnecessarily condescending, infectiously according to his greet-
ings.
Undoubtedly, his attitude toward me had a sobering effect on my business
negotiations during my stay in London.
Today, let me challenge the word problem similar to the prestigious Univer-
sity of Oxford math admission tests that was conducted in 2020.
A cubic has equation \(y = x^{3} + ax^{2} + bx + c\) and has turning points at
\((2, 3)\) and \((4, d)\) for some \(d\). What is the value of \(d\) ?
I changed the turning points of the actual math admission tests at the Uni-
versity of Oxford. Here is my solution to the problem.
The turning points have \(\frac{dy}{dx} = 0\) at \(x = 2\) and at \(x = 4\).
\(y’ = 3x^{2} + 2ax + b\).
So \(12 + 4a + b = 0\), and \(48 + 8a + b = 0\).
Solve these for \(a = -9\), \(b = 24\)
Now \(y = 3\) and \(x = 2\), so \(3 = 8 + (-9)・4 + 24・2 + c\), then \(c = -17\)
Then at \(x = 4\), \(y = 64 + (-9)・16 + 24・4 – 17 = 1\)
\(y = -1\) at \(x = 4\) gives \(d = -1\).
The adjectives, such as courteous, polite, and civil, express well-mannered
attitudes toward others. I will mind my manners, at least being “civil,” like
the taxi driver I first met.
Stay tuned, and expect to see my next post.
Keep well.
Frank Yoshida
 ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄
【グローバルビジネスで役立つ数学】でもっと学習する b^^)
【参考図書】『もう一度高校数学』(著者:高橋一雄氏)株式会社日本実業出版社
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