Question

If f(x)f(x) is an exponential function where f(-0.5)=27f(−0.5)=27 and f(1.5)=21f(1.5)=21, then find the value of f(0.5)f(0.5), to the nearest hundredth.

Answer 1

Given:

f(x) is an exponential function.

$f(-0.5)=27$

$f(1.5)=21$

To find:

The value of f(0.5), to the nearest hundredth.

Solution:

The general exponential function is

$f(x)=ab^x$

For, x=-0.5,

$f(-0.5)=ab^{-0.5}$

$27=ab^{-0.5}$          ...(i)

For, x=1.5,

$f(1.5)=ab^{1.5}$

$21=ab^{1.5}$          ...(ii)

Divide (ii) by (i).

$\dfrac{21}{27}=\dfrac{ab^{1.5}}{ab^{-0.5}}$

$\dfrac{7}{9}=b^2$

Taking square root on both sides, we get

$\dfrac{\sqrt{7}}{3}=b$

$b\approx 0.882$

Putting b=0.882 in (i), we get

$27=a(0.882)^{-0.5}$

$27=a(1.0648)$

$\dfrac{27}{1.0648}=a$

$a\approx 25.357$

Now, the required function is

$f(x)=25.357(0.882)^x$

Putting x=0.5, we get

$f(0.5)=25.357(0.882)^{0.5}$

$f(0.5)=23.81399$

$f(0.5)\approx 23.81$

Therefore, the value of f(0.5) is 23.81.