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If f(a) is an exponential function where f(-3) = 18 and f(1) = 59, then find the
value of f(0), to the nearest hundredth.

Sagot :

Given:

For en exponential function f(a):

[tex]f(-3)=18[/tex]

[tex]f(1)=59[/tex]

To find:

The value of f(0).

Solution:

The general form of an exponential function is:

[tex]f(x)=ab^x[/tex]          ...(i)

Where, a is the initial value and b is the growth/ decay factor.

We have, [tex]f(-3)=18[/tex]. Substitute [tex]x=-3,f(x)=18[/tex] in (i).

[tex]18=ab^{-3}[/tex]            ...(ii)

We have, [tex]f(1)=59[/tex]. Substitute [tex]x=1,f(x)=59[/tex] in (i).

[tex]59=ab^{1}[/tex]            ...(iii)

On dividing (iii) by (ii), we get

[tex]\dfrac{59}{18}=\dfrac{ab^{1}}{ab^{-3}}[/tex]

[tex]3.278=b^{1-(-3)}[/tex]

[tex]3.278=b^{4}[/tex]

[tex](3.278)^{\frac{1}{4}}=b[/tex]

[tex]1.346=b[/tex]

Substituting the value of b in (iii).

[tex]59=a(1.346)^1[/tex]

[tex]\dfrac{59}{1.346}=a[/tex]

[tex]43.83358=a[/tex]

[tex]a\approx 43.83[/tex]

The initial value of the function is 43.83. It means, [tex]f(0)=43.83[/tex].

Therefore, the value of f(0) is 43.83.

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