If $h (x) = 3^{2 x}$ $h(x)=3^(2x)$ and $f (x) = x - 2$ $f(x)=x-2$ , how do you find $h (f (- 3))$ $h(f(-3))$ ?

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Answer 1 · 2018-03-30T05:16:58

$h (f (- 3)) ≅ 0.000017$ $h(f(-3))~=0.000017$

When we evaluate a function, we substitute in a value.

$h (f (x))$ $h(f(x))$ means 'evaluate $f (x)$ $f(x)$ first, then evaluate $h (that answer)$ $h("that answer")$ '.

So first, we know $x = - 3$ $x=-3$

Then $f (x) = f (- 3)$ $f(x)=f(-3)$

We substitute in $- 3$ $-3$ where we see $x$ $x$ , so $f (- 3) = (- 3) - 2 = - 5$ $f(-3)=(-3)-2=-5$

Now we evaluate $h (f (x)) = h (- 5)$ $h(f(x))=h(-5)$ .

$h (x) = 3^{2 x}$ $h(x)=3^(2x)$ and we substitute in $- 5$ $-5$ wherever we see $x$ $x$ :

$h (x) = 3^{2} (- 5) = 3^{- 10} = \frac{1}{3^{10}} ≅ 0.000017$ $h(x)=3^2(-5)=3^-10 =1/3^10~=0.000017$

If h(x)=32xh(x)=3^(2x) and f(x)=x−2f(x)=x-2, how do you find h(f(−3))h(f(-3))?