If h(x)=x2-5+7. h(2)

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Mathematics

1 answer:

Marta_Voda [28]2 years ago

3 0

The way to solve this is by re-writing the equation except with 2 as x.
h(x)= x^2-5+7
h(2)= 2^2-5+7
h(2)=4-5+7
h(2)=-1+7
h(2)= 6

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Suppose h(x)=3x-2 and j(x) = ax +b. Find a relationship between a and b such that h(j(x)) = j(h(x))

sergejj [24]

Answer:

$\displaystyle a = \frac{1}{3} \text{ and } b = \frac{2}{3}$

Step-by-step explanation:

We can use the definition of inverse functions. Recall that if two functions, f and g are inverses, then:

$\displaystyle f(g(x)) = g(f(x)) = x$

So, we can let j be the inverse function of h.

Function h is given by:

$\displaystyle h(x) = y = 3x-2$

Find its inverse. Flip variables:

$x = 3y - 2$

Solve for y. Add:

$\displaystyle x + 2 = 3y$

Hence:

$\displaystyle h^{-1}(x) = j(x) = \frac{x+2}{3} = \frac{1}{3} x + \frac{2}{3}$

Therefore, a = 1/3 and b = 2/3.

We can verify our solution:

$\displaystyle \begin{aligned} h(j(x)) &= h\left( \frac{1}{3} x + \frac{2}{3}\right) \\ \\ &= 3\left(\frac{1}{3}x + \frac{2}{3}\right) -2 \\ \\ &= (x + 2) -2 \\ \\ &= x \end{aligned}$

And:

$\displaystyle \begin{aligned} j(h(x)) &= j\left(3x-2\right) \\ \\ &= \frac{1}{3}\left( 3x-2\right)+\frac{2}{3} \\ \\ &=\left( x- \frac{2}{3}\right) + \frac{2}{3} \\ \\ &= x \stackrel{\checkmark}{=} x\end{aligned}$