Question

A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?

Answer 1

Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

$f(x)=\frac{1}{2^{2x}}$   (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

$\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}]$  (2)

To solve this derivative you use the following derivative formula:

$\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}$

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

$\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)$

You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

$-2(ln2)2^{-2x}=-1$

Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

$-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235$

With this value you calculate f(a):

$f(a)=\frac{1}{2^{2(0.235)}}=0.721$

Next, you use the general equation of line:

$y-y_o=m(x-x_o)$

for xo = a = 0.235 and yo = f(a) = 0.721:

$y-0.721=(-1)(x-0.235)\\\\y=-x+0.956$

The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

$0=-x+0.956\\\\x=0.956$

hence, the x-intercept of the tangent line is 0.956