Question

If f ( x ) f(x) is an exponential function where f ( − 5 ) = 26 f(−5)=26 and f ( 3 ) = 54 f(3)=54, then find the value of f ( 0.5 ) f(0.5), to the nearest hundredth.

Answer 1

Answer:

$f(0.5) = 42.55$

Step-by-step explanation:

Given

$f(-5) = 26$

$f(3) = 54$

Required

Find f(0.5)

An exponential function is of the form:

$y=ab^x$

For: $f(-5) = 26$

$x = -5$    $y = 26$

So, we have:

$26 = ab^{-5}$ --- (1)

For $f(3) = 54$

$x = 3$     $y =54$

So, we have:

$54 = ab^3$ --- (2)

Divide (2) by (1)

$\frac{54}{26} = \frac{ab^3}{ab^{-5}}$

$\frac{54}{26} = \frac{b^3}{b^{-5}}$

Apply law of indices

$\frac{54}{26} = b^{3-(-5)}$

$\frac{54}{26} = b^{3+5}$

$\frac{54}{26} = b^{8}$

$2.07692307692= b^{8}$

Take 8th roots of both sides

$b = \sqrt[8]{2.07692307692}$

$b = 1.09566440144$

$b = 1.10$

Substitute 1.10 for b in $54 = ab^3$

$54 = a * 1.10^3$

$54 = a * 1.331$

Solve for a

$a = \frac{54}{1.331}$

$a = 40.57$

To solve for f(0.5), we have:

$y=ab^x$

$f(0.5) = 40.57 * 1.10 ^ {0.5}$

$f(0.5) = 40.57 * 1.04880884817$

$f(0.5) = 42.5501749703$

$f(0.5) = 42.55$ (approximated)