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I came across this question tutoring someone:

The tangent line to $y = f(x)$ at $x = 2$ has the equation $y = 3 - 7x$.

Find $f(2)$.

My student has only started limits and differentiation. How could you possibly solve this without integration?

Community
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Horse
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3 Answers3

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Hint:

the point $P=(2,f(2))$ is also a point of the tangent line at this point.

Emilio Novati
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Just you can use $$f(2)=3-7\cdot2=-11$$

Michael Rozenberg
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  • Michael, I'm afraid that's the equation of the tangent line. What we are looking for is $f(2)$ instead of $f^{'}(2)$. – Horse Sep 25 '17 at 19:55
  • @Horse $f'(2)=-7$ because it's a slope of $y=-7x+3$. – Michael Rozenberg Sep 25 '17 at 19:57
  • But $f^{'}(x)$ is $3 - 7x$. – Horse Sep 25 '17 at 19:58
  • @Horse $f'(x)$ it's $f'(x)$, but $f'(2)$ it's a slope of the tangent to the graph of $f$ at $x=2$ by definition. Thus, $f'(2)=-7$. – Michael Rozenberg Sep 25 '17 at 20:00
  • Precisely, so $f^{'}(2) = 3 - 7(2) = -11$. What is asked of us is $f(2)$. – Horse Sep 25 '17 at 20:02
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    @Horse No! The touching point it's the common point of graph $f$ and the tangent. Thus, the touching point placed on the tangent, which says that the $y$-coordinate of the touching point we can get so: $y=3-7\cdot2$, which is also $f(2)$. – Michael Rozenberg Sep 25 '17 at 20:06
  • @Horse $f'(x)$ is not the equation of the tangent line. Just look at $f(x) = x$. $f'(x) = 1$. This is not the equation of a line tangent to any point on $f$. – wgrenard Sep 25 '17 at 20:08
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First option

The tangent line to $y=f(x)$ at $x=a$ is

$$y-f(a)=f'(a)(x-a).$$ Thus we have at $x=2$,

$$y=f(2)+f'(2)(x-2)=f'(2)x+f(2)-2f'(2)=-7x+3.$$ That is,

$$\begin{cases}f'(2)&=-7,\\f(2)-2f'(2)&=3.\end{cases}$$

Solve the linear system and you'll get the answer.

Second option

At $x=a$ the tangent line to $f(x)$ and $f(x)$ have the common value $f(a)$. Thus,

$f(2)=3-7\cdot 2=-11.$

mfl
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