Given f(x) = x^2- 3x + 5, what is f(-2)?

Mathematics

2 answers:

vlabodo [156]1 year ago

7 0

Step-by-step explanation:

f(x) = x² - 3x + 5

f(-2) = (-2)² - 3.(-2) + 5

f(-2) = 4 + 6 + 5

f(-2) = 15

Alja [10]1 year ago

7 0

Answer:

Step-by-step explanation:

Comment

In general, f(x) can be thought of as being y. So the equation could read as "Let x = - 2" when y = x^2 - 3x + 5.

Another way of thinking of it is by saying f(-2), wherever you see an x on the right hand side, substitute the value given on the left side (in this case x = - 2

Solution

f(-2) = (-2)^2 - 3x + 5

f(-2) = 4 - 3(-2) + 5 Combine

f(-2) = 4 + 6 + 5 Add

f(-2) = 15

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Find derivative problem<br> Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$