Question

A function f is described by f(x)=A*exp(kx)+B, where A, B and k are constants. Given f(0)=1, f(1)=2, and that the horizontal asymptote of f is -4, the value of k is

Answer 1

Answer:

k = ln (6/5)

Step-by-step explanation:

for

f(x)=A*exp(kx)+B

since f(0)=1, f(1)=2

f(0)= A*exp(k*0)+B = A+B = 1

f(1) = A*exp(k*1)+B =  A*e^k + B = 2

assuming k>0 , the horizontal asymptote H of f(x) is

H= limit f(x) , when x→ (-∞)

when x→ (-∞) , limit f(x) =  limit (A*exp(kx)+B) = A* limit [exp(kx)]+B* limit = A*0 + B = B

since

H= B = (-4)

then

A+B = 1 → A=1-B = 1 -(-4) = 5

then

A*e^k + B = 2

5*e^k + (-4) = 2

k = ln (6/5)    ,

then our assumption is right and k = ln (6/5)  

Answer 2

The value of k is $k=ln(\frac{6}{5} )$.

Given function is,

                           $f(x)=Ae^{kx} +B$

Substitute $f(0)=1,f(1)=2$ in above equation.

We get,

                       $A+B=1\\\\Ae^{k}+B=2$

Given that horizontal asymptote of f is -4.

               $\lim_{x \to -\infty} Ae^{kx}+B=-4\\ \\ B=-4$

So,  $A=1-B=1-(-4)=5$

Substitute value of A and B.

                $5e^{k}-4=2\\ \\e^{k} =\frac{6}{5}\\ \\k=ln(\frac{6}{5} )$

Hence, the value of k is $k=ln(\frac{6}{5} )$.

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