Answer:
The values for the constants are c=2, and d=-4.
Step-by-step explanation:
We have the function f, defined as:
[tex]f(x)=cx+d;x\leq 2[/tex]
and
[tex]f(x)=x^2-cx;x>2[/tex]
where c and d are constants.
So, if f is differentiable at x=2, then it must be continuos at x=2, therefore
[tex]f(2)=\lim_{x \to \ 2^+} f(x) \Leftrightarrow2c+d=4-2c\Leftrightarrow d=4-4c[/tex]
Let's call this, equation 1.
On the other hand, we have that f is differentiable, therefore
[tex]f'(2)= \lim_{x \to \ 2^+} f'(x)[/tex]
and
[tex]f'(x)=c;x\leq 2[/tex]
[tex]f'(x)=2x-c;x>2[/tex]
so, we calculate
[tex]f'(2)= \lim_{x \to \ 2^+} f'(x)\Leftrightarrow c=4-c\Leftrightarrow c=2[/tex]
which, replacing in equation 1, gives us
[tex]d=4-4c=4-8\Leftrightarrow d=-4[/tex]
Finally, the answer is that c=2, and d=-4.
A function that is differentiable at a given value, will be continuous at that value.
The values of c and d are 2 and -4, respectively.
The function is given as:
[tex]\mathbf{f(x) = cx + d,\ x \le 2}[/tex]
[tex]\mathbf{f(x) = x^2 - cx,\ x > 2}[/tex]
Recall that a function will be continuous and differentiable at the same value.
This means that:
[tex]\mathbf{f(2) = 2c + d}[/tex]
and
[tex]\mathbf{f(2) = 2^2 - 2c}[/tex]
[tex]\mathbf{f(2) = 4 - 2c}[/tex]
So, we have:
[tex]\mathbf{2c + d = 4 - 2c}[/tex]
Collect like terms
[tex]\mathbf{d = 4 - 4c}[/tex]
Next, calculate f'(x)
[tex]\mathbf{f'(x) =c}[/tex]
[tex]\mathbf{f'(x) =2x - c}[/tex]
The function is differentiable at x = 2
So:
[tex]\mathbf{f'(2) = c}[/tex]
[tex]\mathbf{f'(2) =2\times 2 - c = 4-c}[/tex]
So, we have:
[tex]\mathbf{c = 4-c}[/tex]
Add c to both sides
[tex]\mathbf{2c = 4}[/tex]
Divide both sides by 2
[tex]\mathbf{c = 2}[/tex]
Substitute 2 for c in [tex]\mathbf{d = 4 - 4c}[/tex]
[tex]\mathbf{d = 4 -4(2)}[/tex]
[tex]\mathbf{d = -4}[/tex]
Hence, the values of c and d are 2 and -4, respectively.
Read more about differentiable and continuous functions at:
https://brainly.com/question/15667266
