WILL GIVE BRAINLY FOR ANSWER!! Please help with this question!!! Given the piecewise function: f(x) = 1/2x + 5, x > 2 6, x = 2 x + 4, x < 2 a. Write f' (f prime) as a piecewise function b. Determine if f is differentiable at x = 2. Give a reason for your answer. Photo is attatched.

Answer:

A)

[tex]f'(x) = \left\{ \begin{array}{lIl} \frac{1}{2} & \quad x >2 \\ 0& \quad x =2\\1&\quad x<2 \end{array} \right.[/tex]

B) Continuous but not differentiable.

Step-by-step explanation:

So we have the piecewise function:

[tex]f(x) = \left\{ \begin{array}{lIl} \frac{1}{2}x+5 & \quad x >2 \\ 6& \quad x =2\\x +4&\quad x<2 \end{array} \right.[/tex]

A)

To write the differentiated piecewise function, let's differentiate each equation separately. Thus:

1)

[tex]\frac{d}{dx}[\frac{1}{2}x+5}][/tex]

Expand:

[tex]\frac{d}{dx}[\frac{1}{2}x]+\frac{d}{dx}[5][/tex]

The derivative of a linear equation is just the slope. The derivative of a constant is 0. Thus:

[tex]\frac{d}{dx}[\frac{1}{2}x+5}]=\frac{1}{2}[/tex]

2)

[tex]\frac{d}{dx}[6][/tex]

Again, the derivative of a constant is 0. Thus:

[tex]\frac{d}{dx}[6]=0[/tex]

3)

We have:

[tex]\frac{d}{dx}[x+4][/tex]

Expand:

[tex]\frac{d}{dx}[x]+\frac{d}{dx}[4][/tex]

Simplify:

[tex]=1[/tex]

Now, let's substitute our original equations for the differentiated equations. The inequalities will stay the same. Therefore:

[tex]f'(x) = \left\{ \begin{array}{lIl} \frac{1}{2} & \quad x >2 \\ 0& \quad x =2\\1&\quad x<2 \end{array} \right.[/tex]

B)

For a function to be differentiable at a point, the function must be a) continuous at that point, and b) the left and right hand derivatives must be equivalent.

Let's first determine if the function is continuous at the point. Remember that a function is continuous at a point if and only if:

[tex]\lim_{x \to n^-} f(n)= \lim_{x \to n^+}f(n)=f(n)[/tex]

Let's find the left hand limit of f(x) at it approaches 2.

[tex]\lim_{x \to 2^-}f(x)[/tex]

Since it's coming from the left, let's use the third equation:

[tex]\lim_{x \to 2^-}f(x)\\=\lim_{x \to 2^-}(x+4)[/tex]

Direct substitution:

[tex]=(2+4)=6[/tex]

So:

[tex]\lim_{x \to 2^-}f(x)=6[/tex]

Now, let's find the right-hand limit:

[tex]\lim_{x \to 2^+}f(x)[/tex]

Since we're coming from the right, let's use the first equation:

[tex]\lim_{x \to 2^+}(\frac{1}{2}x+5)[/tex]

Direct substitution:

[tex](\frac{1}{2}(2)+5)[/tex]

Multiply and add:

[tex]=6[/tex]

So, both the left and right hand limits are equivalent. Now, find the limit at x=2.  

From the piecewise function, we can see that the value of f(2) is 6.

Therefore, the function is continuous at x=2.

Now, let's determine differentiability at x=2.

For a function to be differentiable at a point, both the right hand and left hand derivatives must be equivalent.

So, let's find the derivative of the function as x approahces 2 from the left and from the right.

From the differentiated piecewise function, we can see that as x approaches 2 from the left, the derivative is 1.

As x approaches 2 from the right, the derivative is 1/2.

Therefore, the right and left hand derivatives are not the same.

Thus, the function is continuous but not differentiable.

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