Logarithm 9

Calculus Level 3

$\large f(x) = (x - \pi)(x - 2\pi)(x - \pi^{2})$

For $f(x)$ as given above, find $f'(x_1)$ , where $x_1 = e^{\ln (\pi^{2}) + \ln (\pi^{3}) - \frac{\log (\pi^{4})}{\log (e)}}$ .

Notations:

$f '(x)$ the first derivative of $f (x)$ .
$\ln(\cdot)$ denotes the natural logarithm.
$\log (\cdot)$ denotes the common logarithm.
$\pi \approx 3.14159$ is the ratio between a circle's circumference and diameter.

f '(x_1) = \pi

f '(x_1) < \pi

f '(x_1) > \pi^{2}

f '(x_1) = \pi^{2}

\pi < f '(x_1) < \pi^{2}

1 solution

Chew-Seong Cheong
Apr 10, 2017

Note that:

$\begin{aligned} x_1 & = \exp \left(\ln \pi^2 + \ln \pi^3 - \frac {\log_a \pi^4}{\log_a e} \right) \\ & = \exp \left(\ln \pi^2 + \ln \pi^3 - \frac {\frac {\ln \pi^4}{\ln a}}{\frac {\ln e}{\ln a}} \right) \\ & = \exp \left(\ln \pi^2 + \ln \pi^3 - \frac {\ln \pi^4}{\ln e} \right) \\ & = \exp \left(2\ln \pi + 3\ln \pi - \frac {4\ln \pi}{1} \right) \\ & = e^{\ln \pi} = \pi \end{aligned}$

$\begin{aligned} f(x) & = (x - \pi)(x - 2\pi)(x - \pi^2) \\ f'(x) & = (x - 2\pi)(x - \pi^2) + (x - \pi)(x - \pi^2) + (x - \pi)(x - 2\pi) \\ f'(x_1) & = \require {cancel} (\pi - 2\pi)(\pi - \pi^2) + \cancel {(\pi - \pi)}^0(x - \pi^2) + \cancel {(\pi - \pi)}^0(x - 2\pi) \\ & = \pi^2 (\pi -1) \\ & \boxed{> \pi^2} \end{aligned}$

Logarithm 9

1 solution

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