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ollegr [7]
3 years ago
8

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.

Mathematics
1 answer:
Katarina [22]3 years ago
3 0

Answer:

A)

f'(x) = \left\{        \begin{array}{lIl}            \frac{1}{2} & \quad x >2 \\            0& \quad x =2\\1&\quad x

B) Continuous but not differentiable.

Step-by-step explanation:

So we have the piecewise function:

f(x) = \left\{        \begin{array}{lIl}            \frac{1}{2}x+5 & \quad x >2 \\            6& \quad x =2\\x +4&\quad x

A)

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

1)

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

Expand:

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

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

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

2)

\frac{d}{dx}[6]

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

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

3)

We have:

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

Expand:

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

Simplify:

=1

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

f'(x) = \left\{        \begin{array}{lIl}            \frac{1}{2} & \quad x >2 \\            0& \quad x =2\\1&\quad x

B)

For a function to be differentiable at a point, the function <em>must </em>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:

\lim_{x \to n^-} f(n)= \lim_{x \to n^+}f(n)=f(n)

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

\lim_{x \to 2^-}f(x)

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

\lim_{x \to 2^-}f(x)\\=\lim_{x \to 2^-}(x+4)

Direct substitution:

=(2+4)=6

So:

\lim_{x \to 2^-}f(x)=6

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

\lim_{x \to 2^+}f(x)

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

\lim_{x \to 2^+}(\frac{1}{2}x+5)

Direct substitution:

(\frac{1}{2}(2)+5)

Multiply and add:

=6

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 <em>not</em> the same.

Thus, the function is continuous but <em>not</em> differentiable.

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basically 'because it represents the solution so we solved it and it is correct'


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