The function #f(x) = tan(3^x)# has one zero in the interval #[0, 1.4]#. What is the derivative at this point?

Question

The function #f(x) = tan(3^x)# has one zero in the interval #[0, 1.4]#. What is the derivative at this point?

Cameron Alexander · Accepted Answer

#pi ln3#

If #tan(3^x) = 0#, then #sin(3^x) = 0# and #cos(3^x) = +-1#

Therefore #3^x# = #kpi# for some integer #k#.

We were told that there is one zero on #[0,1.4]#. That zero is NOT #x=0# (since #tan 1 != 0#). The smallest positive solution must have #3^x = pi#.

Hence, #x = log_3 pi#.

Now let's look at the derivative.

#f'(x) = sec^2(3^x) * 3^x ln3#

We know from above that #3^x = pi#, so at that point

#f' = sec^2(pi) * pi ln3 =(-1)^2 pi ln3 = pi ln3#

Charles Anctil · Answer

undefined To find the derivative of $ f(x) = \tan(3^x) $ at the point where it has a zero in the interval [0, 1.4], we need to find the derivative of $ f(x) $ and then evaluate it at the zero.

First, find the derivative $ f'(x) $:

$$ f'(x) = \frac{d}{dx} \tan(3^x) $$

Using the chain rule for differentiation, the derivative is:

$$ f'(x) = \sec^2(3^x) \cdot 3^x \ln(3) $$

Now, to find the zero of $ f(x) $ in the interval [0, 1.4], we need to solve:

$$ \tan(3^x) = 0 $$

This occurs when $ 3^x $ is an odd multiple of $ \frac{\pi}{2} $. Since $ \pi $ is approximately 3.14159, an odd multiple of $ \frac{\pi}{2} $ will be $ \frac{\pi}{2} $, $ \frac{3\pi}{2} $, etc.

Let's solve for $ x $ using $ 3^x = \frac{\pi}{2} $:

$$ x = \log_3\left(\frac{\pi}{2}\right) $$

Evaluate $ f'(x) $ at this $ x $:

$$ f'\left(\log_3\left(\frac{\pi}{2}\right)\right) = \sec^2\left(\frac{\pi}{2}\right) \cdot \frac{\pi}{2} \ln(3) $$

$$ \sec^2\left(\frac{\pi}{2}\right) = 1 $$ because $ \sec(x) $ is the reciprocal of $ \cos(x) $ and $ \cos\left(\frac{\pi}{2}\right) = 0 $.

Thus, the derivative $ f'(x) $ at the point where $ f(x) $ has a zero in the interval [0, 1.4] is:

$$ f'\left(\log_3\left(\frac{\pi}{2}\right)\right) = \frac{\pi}{2} \ln(3) $$

Charles Arocha · Answer

undefined To find the derivative of $ f(x) = \tan(3^x) $ at the point where it crosses the x-axis (i.e., where $ f(x) = 0 $), we first need to find that point.

Given $ f(x) = \tan(3^x) $, we set $ f(x) = 0 $:

$$ \tan(3^x) = 0 $$

The solutions to this equation occur when $ \tan(3^x) $ equals zero. The tangent function is zero at multiples of $ \pi $. Therefore, we can set up the equation:

$$ 3^x = n\pi $$
Where $ n $ is an integer.

Now, we solve for $ x $ in the interval [0, 1.4]:

$$ 3^x = n\pi $$

Taking the natural logarithm of both sides:

$$ x \ln(3) = \ln(n\pi) $$

$$ x = \frac{\ln(n\pi)}{\ln(3)} $$

By trying different integer values for $ n $ within the interval [0, 1.4], we can find the appropriate value of $ n $. Note that only one value of $ n $ will yield a solution within the specified interval.

After finding the value of $ x $, we can then find the derivative $ f'(x) $ using the chain rule:

$$ f'(x) = \sec^2(3^x) \cdot 3^x \ln(3) $$

Evaluate $ f'(x) $ at the found $ x $ value to get the derivative at that point.