I’m in IB Math and we are working on some calculus problems but I wanted to get extra practice so this is a problem in my book. The number in parenthesis next to the parts are the “marks” we get for the question if we get it right. So, usually that’s about how much work we have to show or how many steps to take.

Let $f(x) = \ln x - 5x$, for $x > 0$.

a) Find $f’(x)$. (2)

Would this be $1/x - 5$?

b) Find $f’’(x)$. (1)

Would this be simply $1/x$?

c) Solve for $f’(x) = f’’(x)$. (2)

We set them up equal to each other. So,

$1/x - 5$ = $1/x$

The $1/x$’s cancel out and we get $-5$. Is this correct?

  • 2
    $\begingroup$ The derivative of $\frac 1x$ is not $\frac 1x$. $\endgroup$
    – lulu
    Dec 13, 2018 at 15:30

3 Answers 3


$a)$ Correct.

$b)$ Incorrect. The derivative of $\frac{1}{x}$ is not $\frac{1}{x}$. You can use the Power Rule.

$$f’(x) = \frac{1}{x}-5= x^{-1}-5$$

$$f’’(x) = -1x^{-2} = -x^{-2} = -\frac{1}{x^2}$$

$c)$ Let

$$f’(x) = f’’(x)$$

which results in

$$\frac{1}{x}-5 = -\frac{1}{x^2}$$

$$x^2\bigg(\frac{1}{x}-5 = -\frac{1}{x^2}\bigg)$$

$$x-5x^2 = -1$$

$$5x^2-x-1 = 0$$

Using the Quadratic Formula yields

$$x = \frac{-(-1)\pm\sqrt{(-1)^2-4(5)(-1)}}{2(5)}$$

$$x = \frac{1\pm\sqrt{21}}{10}$$


$a)$ Yes

$b)$ No. $$f'(x)=1/x-5$$

$$\implies f''(x)=-1/x^2$$

because $$\frac{d}{dx}(1/x)=-1/x^2$$

$c)$ $$-1/x^2=1/x-5$$


$$\implies x=\frac{1+\sqrt{21}}{10}\quad \text{or}\quad \frac{1-\sqrt{21}}{10}$$


Point "a)" is ok.

For point "b)" recall that

$$\frac1x = x^{-1} \implies \frac{d}{dx}(x^{-1})=-x^{-2}=-\frac1{x^2}$$

Note also that for point "c)" in nay case

$$\frac1x - 5 = \frac1x \iff -5=0$$

that would mean that the equation has not solutions.


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