The function f(x)= tan(3^x) has one zero in the interval [0, 1.4]. The derivative at this point is:
a. 0.411
b. 1.042
c. 3.451
d. 3.763
e. undefined
5 Answers

The tangent function has zeros at every k*Pi where k is an integer. So to find the zeros of this function, we would need to check the values of x that make 3^x = k*Pi for some integer k.
k < 0: Since 3^x is always positive, no negative values of k would work.
k = 0: Notice that 3^x is never zero!
k > 0: The only logical choice for k would be k = 1 since we need to make sure x stays in the interval [0, 1.4].
So let’s suppose k = 1.
Then 3^x = Pi => x = log Pi (where log is in base 3)
(Its value is about 1.04 which is in [0, 1.4] as needed)
So now we have found where f(x) has a zero (i.e. at x = log Pi where log is in base 3).
Taking the derivative, we get:
f'(x) = [sec(3^x)]^2 * 3^x * ln 3 (where ln = log in base e)
Plugging in x = log Pi (where log is in base 3) into f'(x), we get:
f'(log Pi)
= [sec(3^(log Pi))]^2 * 3^(log Pi) * ln 3
= [sec(Pi)]^2 * Pi * ln 3
= (1)^2 * Pi * ln 3
= Pi * ln 3

wager it relies upon on the prism one seems by using; I observed extra of a bumbling Key Stone Cop overall performance of the Washington Nationals than I did of the Cincinnati Reds “super” overall performance. If i became prepared on the Reds or a fan on the ballpark i might have enjoyed it and that i might have stayed on the park until the tip of the sport. The ninety 3 12 months previous WW II Veteran I watched the sport with begged me to alter the sport whilst it became 80; i ultimately did whilst it became 100. immediately is a clean interest and that i nevertheless say the Nationals take this series 2a million.

Kind of confused on what you’re asking. First you are describing where one should look at the graph (the interval) of ytan(3^x) and then you want the derivative, but you never specify at what point. The point (0,1.4) does not exist on the graph y=tan(3^x)..
If you wan to restate your problem, you can email me, and I’ll try to figure it out.

The range of f(x) for x=0 until 1.4 is: tan(1)< f(x) < tan(4.66); thus f(x)=0 is false; thus no point to answer this question.
Oh, sorry! 3^1.4 = 4.66 < 3pi/2, while 3^1 < pi;
and tan(pi)= 0 =tan(3^x); so x0=1.042;
y’=(ln3)*3^x /(cos(3^x))^2, y(x0) = 1.099*pi /(cos(pi))^2 = 3.45;

e^u = 3^x
u = xln3
du = ln3dx
f'(x) = ln3(3^x)sec^2^3^x)
3^x = arctan(0)
3^x = 0, π
xln3 = lnπ
x = lnπ/ln3 = 1.0420
f'(x) = ln3(3^π/ln3)sec^2^3^π/ln3)
f'(x) = ln3(3^1.0420)sec^2(3^1.0420)
f'(x) = ln3(3.14159)sec^2(3.14159))
f'(x) = 3.4513