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?

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

  (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):

 (2)

To solve this derivative you use the following derivative formula:

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

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.

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

With this value you calculate f(a):

Next, you use the general equation of line:

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

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:

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

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval