Question

Calculus help please!

Answer 1

@Vocaloid

Answer 2

@Vocaloid please help

Answer 3

1 is e

Answer 4

@Mercury

Answer 5

uhtook me a while but I think I got the same thing

Answer 6

d for the other one?

Answer 7

this one b?

Answer 8

yeah that's what i got too

Answer 9

help some more?

Answer 10

1 is d?

Answer 11

well if the derivative is 1/2 then the slope of the tangent line is also 1/2. since it's asking for the normal (aka perpendicular) the final slope must be the negative reciprocal -2

Answer 12

yay and 2 e I think

Answer 13

if the slope of the normal is -2 then D cannot be the answer.

Answer 14

oh okay

Answer 15

second one is also wrong hold on

Answer 16

oh wait I see, you have to plug in |cos(ax)|, try each a-value one by one, and see which is the smallest a-value that still makes |cos(ax)| differentiable from 0 to 3

Answer 17

so you could start with B, |cos(0.3x)| and see if there the function is still differentiable across the entire interval 0 to 3 (no sharp edges) and if not, go up to the next higher a value

Answer 18

just use software to graph them and inspect visually for any sharp edges

Answer 19

so I'm confused is d the answer of no for 1 ?or is it a

Answer 20

it has to be A) since it's the only one with slope -2

if the slope of the tangent is 1/2 then the slope of the normal has to be -2

Answer 21

okay!

Answer 22

ok, it asks for a and b values that will make it differentiable across ALL values of x

so you'd need to find which a value and b value makes 2x + 1 = ax^2 + b always true

Answer 23

idk i think i'm confusing myself with this one >>

Answer 24

Your [piece-wise] function is:

\(f(x) = \begin{cases}
2x+1 & x\ \le 1\\
ax^2 + b & x > 1
\end{cases} \)

Meaning that:


\(f'(x) = \begin{cases}
2 & x\ \le 1\\
2ax & x > 1
\end{cases} \)

So we have differentiated, but we want to know the [piece-wise] function, with actual values for \(a\) and \(b\) so that, the function is differentiable.

Skipping some jargon, that means that it should have the same value and slope from both functions at \(x = 1\).

IOW, if we say that \( f_1(x) = 2x+1\) and \(f_2(x) = ax^2 + b\), then we need to show this:

\(\begin{cases}
{value } & f_1(1) = f_2(1)\\
{slope} & f'_1(1) = f'_2(1)
\end{cases} \)

So we end up with this algebra:

\( \begin{cases}
\ {value} & 2(1) + 1 = a(1)^2 + b \\
{slope} & 2 = 2a(1)
\end{cases} \)

Or:

\( \begin{cases}
\ 3 = a + b \\
1 = a
\end{cases} \)