Answer:
x-intercept = 0.956
Step-by-step explanation:
You have the function f(x) given by:
[tex]f(x)=\frac{1}{2^{2x}}[/tex] (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):
[tex]\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}][/tex] (2)
To solve this derivative you use the following derivative formula:
[tex]\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}[/tex]
For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:
[tex]\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)[/tex]
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.
[tex]-2(ln2)2^{-2x}=-1[/tex]
Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:
[tex]-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[/tex]
With this value you calculate f(a):
[tex]f(a)=\frac{1}{2^{2(0.235)}}=0.721[/tex]
Next, you use the general equation of line:
[tex]y-y_o=m(x-x_o)[/tex]
for xo = a = 0.235 and yo = f(a) = 0.721:
[tex]y-0.721=(-1)(x-0.235)\\\\y=-x+0.956[/tex]
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:
[tex]0=-x+0.956\\\\x=0.956[/tex]
hence, the x-intercept of the tangent line is 0.956
