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2 years ago

Find the equation of the line tangent to the graph of f(x)=(lnx)^(4)at x=10

Mathematics

1 answer:

Diano4ka-milaya [45]2 years ago

4 0

Hello,

Step-by-step explanation:

$f(x) = ln(x) {}^{4}$

$(ln(u)') = \frac{u'}{u}$

$f'(x) = \frac{4ln {}^{} (x) {}^{3} }{x}$

$f'(10) = \frac{4ln {}^{} (10) {}^{3} }{10} = \frac{12ln(x)}{x}$

$f(10) = ln(10) {}^{4}$

$y = \frac{12ln(x)}{x} (x - 10) + 4ln(10)$

$y = f'(a)(x - a) + f(a)$

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Step-by-step explanation:

The tangent function $y = \tan(x)$ has a real value for real inputs $x$ as long as the input $x \ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

Hence, the domain of the original tangent function is $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

On the other hand, in the function $f(x) = 3\, \tan(23\, x)$ , the input to the tangent function is replaced with $(23\, x)$ .

The transformed tangent function $\tan(23\, x)$ would have a real value as long as its input $(23\, x)$ ensures that $23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

In other words, $\tan(23\, x)$ would have a real value as long as $x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right)$ .

Accordingly, the domain of $f(x) = 3\, \tan(23\, x)$ would be $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

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