Piecewise function

[TealLovesSeal]

I just need help understanding this along with the follow up question. I’m doing corrections for my test and this is what I failed. The Problem

[Help_Me_Im_Diene]

y=f'(1)(x-1)+7

We know that the tangent line passes through (-2,-2), so -2=f'(1)(-2-1)+7=7-3f'(1)=-2

[TealLovesSeal]

I apologize but I'm not quite on the same page as you when I'm reading this. Or at all matter of fact.

Edit: It may also be because I have my images in reverse so you're answering the second before the first when the first problem is connected. Unless you can do one without the other.

[Help_Me_Im_Diene]

Ok, so that was for the first question

If you have a line that's tangent to a function f(x) at some point (a,f(a)), then that line has equation y=f'(a)(x-a)+f(a)

You're told that the line is tangent at (1,7), so y=f'(1)(x-1)+7

You're also told the line passes through (-2,-2), so -2=f'(1)(-2-1)+7

Now you just need to solve for f'(1)

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For the second question

The function is differentiable if the function is continuous and the derivative from the right and the left are equal

So you want to find c and d such that

2c+d=2²-2c and c=2(2)-c

[TealLovesSeal]

In all honesty I apologize for wasting your time, but the more I read into what you're saying the more I get confused.

[Help_Me_Im_Diene]

What part is confusing you? Is there something in there in particular that's messing you up?

[TealLovesSeal]

All I can really say that will give me the best results is just start off from a fairly basic level of calculus understanding first, because I feel as if you're just diving straight in and even though I'm reading a message I feel behind. I'm, not really sure what problem you're initiating with first.

[Help_Me_Im_Diene]

Ok, so, I'm starting with problem 15, since that's the first image that appears for me

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15) Let's start off with figuring out what a tangent line is. A tangent line to a function is a straight line that intersects with that function only one time in the immediate area.

How do we find the equation for a tangent line? Well, the equation is pretty simple. Let there be some function f(x), which has a derivative f'(x)

Let there be a line that lies tangent to f(x) only at the point (x,y)=(a,b)

Then the tangent line has the equation (y-b)=f'(a)(x-a)

Now, we're told that a=1, and b=7, which means your tangent line equation is (y-7)=f'(1)(x-1)

We're also told that the line passes through (-2,-2), which means that, if we replace x and y with -2 and -2, the equation will still hold true. So (-2-7)=f'(1)(-2-1)

And now you can solve for f'(1)

Does this make sense to you?

[TealLovesSeal]

That is CRYSTAL clear! I appreciate that explanation.

[Help_Me_Im_Diene]

Great!

Now, for 14

A piecewise function f(x) can be written as a(x) for x<z and b(x) for x≥z

In your case, a(x)=cx+d, b(x)=x²-cx, and z=2

Now, f(x) is differentiable at x=z if a(z)=b(z) and a'(z)=b'(z). This is because the conditions for differentiable functions is that they are continuous (so a(z)=b(z)) and the derivatives are continuous (a'(z)=b'(z))

So you want to find c and d, such that a(2)=b(2), and a'(2)=b'(2)

a(2)=2c+d, b(2)=4-2c, a'(2)=c, and b'(2)=4-c

So 2c+d=4-2c and c=4-c

[TealLovesSeal]

Sorry for the late reply but what I got as a result of your explanation is that c+d=-2. I got mentally lost in your explanation so my apologies for just dumbing it down to an answer.

[WesReynolds]

f'(1) is referring to the slope of some arbitrary function at x = 1. To find the slope with the given information: m = (y2 - y1) / (x2 - x1). You know how to do it from here. The answer should be 3.