Find a, b and c to make piecewise function f differentiable at x=-1

mathsgreedy

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for
ax^2 + bx if x<-1,
3 if x=-1,
x^3 -cx if x>-1.

for what values of a, b and c will f be differentiable at x=-1?

I understand that I must equate the limits and have found that c must equal 4. However, I am completely lost for ideas on how to find a and b because they seem to cancel out and not give an answer so I must be doing something incorrectly. I have been working on this for days for an assignment due tomorrow and am out of ideas. Any help would be appreciated. Thank you in advance.
 
for
ax^2 + bx if x<-1,
3 if x=-1,
x^3 -cx if x>-1.

for what values of a, b and c will f be differentiable at x=-1?

I understand that I must equate the limits and have found that c must equal 4. However, I am completely lost for ideas on how to find a and b because they seem to cancel out and not give an answer so I must be doing something incorrectly. I have been working on this for days for an assignment due tomorrow and am out of ideas. Any help would be appreciated. Thank you in advance.
Did you calculate slope of the function at at x = -1?
 
I understand that I must equate the limits and have found that c must equal 4.
"Equating limits" would give a- b= c, not c= 4. Using the fact that c must be continuous at x= -1 and equating to limit from the right to the value, 3, at x= -1 gives c= 4. But equating the limit from the left to 3 gives a- b= 3. Now use the fact that \(\displaystyle \lim_{h\to 0^-}\frac{f(-1+ h)- 3}{h}= \lim_{h\to 0^+}\frac{f(-1+ h)- 3}{h}\) to get another equation for a and b.
 
for
ax^2 + bx if x<-1,
3 if x=-1,
x^3 -cx if x>-1.

for what values of a, b and c will f be differentiable at x=-1?

I understand that I must equate the limits and have found that c must equal 4. However, I am completely lost for ideas on how to find a and b because they seem to cancel out and not give an answer so I must be doing something incorrectly. I have been working on this for days for an assignment due tomorrow and am out of ideas. Any help would be appreciated. Thank you in advance.
What you need to remember for this problem is that

If a function is differentiable over an interval, it is continuous over that interval. However, if a function is continuous over an interval, it may or may not be differentiable over that interval. In other words

\(\displaystyle \text {differentiability }\implies \text { continuity BUT continuity does NOT } \implies \text { differentiability.}\)

Now let's go to your work.

To make the function continuous at - 1 we require:

\(\displaystyle \displaystyle \lim_{x \rightarrow -1^-} ax^2 + bx = f(-\ 1) = \lim_{x \rightarrow -1^+}x^3 - cx.\)

Solving those we get

\(\displaystyle (-\ 1)^3 - c(-\ 1) = 3 \implies c - 1 = 3 \implies c = 4.\)

So you were fine to here.

We also have

\(\displaystyle a(-\ 1)^2 + b(-\ 1) = 3 \implies a - b = 3 \implies a = 3 + b.\)

They do NOT cancel out. What you have is an infinite number of possible pairs. For example.

\(\displaystyle a = 19,\ b = 16 \implies 19(-\ 1)^2 + 16(-\ 1) = 19 - 16 = 3.\)

\(\displaystyle a = \dfrac{-\ \sqrt{43}}{7},\ b = \dfrac{-\ \sqrt{43} - 21}{7} \implies \)

\(\displaystyle \left ( \dfrac{-\ \sqrt{43}}{7} \right ) * (-\ 1)^2 + \left ( \dfrac{-\ \sqrt{43} - 21}{7} \right ) * (-\ 1) = \dfrac{-\ \sqrt{43}}{7} + \dfrac{\sqrt{43} + 21}{7} = \dfrac{21}{7} = 3.\)

But you have not considered the issue of differentiability yet. That allows you to distinguish between the possible pairs.

To be differentiable requires that the derivative, which is a LIMIT, exists.

What does that mean in practical terms with respect to f'(- 1)?
 
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Thank you very much! This explained everything really clearly, my problem was distinguishing between continuity and differentiability (sigh, beginner mistakes). I understand this very well now and got a=-2 and b=-5 which seems quite reasonable when I check the graph of the resulting function! Thank you to everyone else as well for helping to guide me through this, very much appreciated!
 
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