tan(3^x) has one zero in the interval [0, 1.4]?

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

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• Kind of confused on what you're asking. First you are describing where one should look at the graph (the interval) of y-tan(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