Question

If f(x) is an exponential function where f(-2) = 1 and f(7) = 63, then find the
value of f(1), to the nearest hundredth.

Answer 1

An exponential function is represented as: $\mathbf{f(x) = ab^x}$

The value of f(1) is 3.95

f(-2) = 1 means that:

$\mathbf{ab^{-2} = 1}$

f(7) = 63 means that:

$\mathbf{ab^{7} = 63}$

Divide both equations

$\mathbf{\frac{ab^7}{ab^{-2}} = \frac{63}{1}}$

Apply law of indices

$\mathbf{b^9 = 63}$

Take 9th root of both sides

$\mathbf{b =1.58}$

Make a, the subject in $\mathbf{ab^{-2} = 1}$

$\mathbf{a = \frac{1}{b^{-2}}}$

$\mathbf{a = b^2}$

Substitute $\mathbf{b =1.58}$

$\mathbf{a = 1.58^2}$

$\mathbf{a = 2.50}$

So, we have:

$\mathbf{f(x) = ab^x}$

$\mathbf{f(x) = 2.50 \times 1.58^x}$

Substitute 1 for x

$\mathbf{f(1) = 2.50 \times 1.58^1}$

$\mathbf{f(1) = 3.95}$

Hence, the value of f(1) is 3.95

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