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I'm not sure how to solve this. $$ f'(5) = \lim_{h\to 0} \frac{2(6+h)^2 - 72}{h} $$ It is $f$ prime of $5$ above. Solve for f(x). If someone could please explain the process of solving and not just give an answer, that would be most helpful.

Thank you!

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  • $\begingroup$ There are very many different functions with $f'(5)=24$, which is all your equation claims. $\endgroup$ Commented Jul 12, 2014 at 17:56
  • $\begingroup$ I just edited it. It is solve for f(x). $\endgroup$
    – androidguy
    Commented Jul 12, 2014 at 17:57
  • $\begingroup$ x @android: You can't "solve for $f$" when all you know about $f$ is that $f'(5)=24$. One possibility would be $f(x)=24x$. Or $f(x)=24x+42\pi$. Or $f(x)=\frac{24}{e^5}e^x$. Or $f(x)=\sin(24x-120)$. $\endgroup$ Commented Jul 12, 2014 at 18:00
  • $\begingroup$ @androidguy Henning is correct, the problem is not well posed. However, if we assume you mean that $\frac{2(6+h)^2-72}{h} \equiv \frac{f(5+h)-f(5)}{h}$ then we can solve it uniquely (or trivially I should say) ($f(x) = 2(1+x)^2$). $\endgroup$
    – Winther
    Commented Jul 12, 2014 at 18:05

2 Answers 2

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It is the derivative of $f(x) = 2(x+1)^2$ at $x = 5$. Thus:

$f(5+h) = 2(6+h)^2$, and $f(5) = 2\cdot 6^2 = 72$. Thus:

$f'(5) = \displaystyle \lim_{h \to 0} \dfrac{f(5+h) - f(5)}{h} = \displaystyle \lim_{h \to 0} \dfrac{2(6+h)^2 - 72}{h} = .... = 24$.

My original answer was that $f(x) = 2x^2$ at $x = 6$. Both gives the same answer $24$, but of course "the right function" is $f(x) = 2(x+1)^2$.

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  • $\begingroup$ Or it could be the derivative of $f(x) = 2(x+1)^2$ at $x = 5$. $\endgroup$ Commented Jul 12, 2014 at 17:59
  • $\begingroup$ Okay, thanks for the help! $\endgroup$
    – androidguy
    Commented Jul 12, 2014 at 18:07
  • $\begingroup$ So for f(x) = 2(x + 1)^2 I could just say x = 5? $\endgroup$
    – androidguy
    Commented Jul 12, 2014 at 18:12
  • $\begingroup$ Yes, you could. $\endgroup$
    – DeepSea
    Commented Jul 12, 2014 at 20:05
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We obviously want to morph that limit into something of the form

\begin{equation} \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} \end{equation}

Luckily, in the question you provided, everything is already in the correct form - that is, you can easily see that $f(x)$ is something like $2(x)^2$.

If we were to use that back-of-the-envelope guess and put it in the limit-equation that I first wrote, we would have

$$\lim_{h\rightarrow0} \frac{2(x+h)^2-2x^2}{h}$$

And if you were to plug in $x=6$, it would essentially be what you wrote in your original question. However, that would correspond to $f'(6)$, and we want it to be $f'(5)$. To fix that, simply change $f(x)$ to $2(x+1)^2$. That would then change the limit-equation to

$$\lim_{h\rightarrow0} \frac{2(x+1+h)^2-2(x+1)^2}{h}$$

Setting $x=5$, we arrive at

$$\lim_{h\rightarrow0} \frac{2(6+h)^2-2(6)^2}{h} = \lim_{h\rightarrow0} \frac{2(6+h)^2-72}{h} = f'(5)$$

So to answer you primary question,

$$\boxed{f(x) = 2(x+1)^2}$$

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  • $\begingroup$ Thanks, this was really helpful! $\endgroup$
    – androidguy
    Commented Jul 12, 2014 at 18:22

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