Question

4 mark for review
$f(x) = \

$$\begin{cases} 3^x & \\text{for } 0 < x < 1 \\\\\\frac{1}{2}x^2 - x + \\frac{7}{2} & \\text{for } 1 < x < 2 \\end{cases}$$

$
let $f$ be the function defined above. which of the following statements is true?

Explanation:

Since the problem is about a piece - wise function and we need to analyze its properties (probably continuity or differentiability at \(x = 1\), but the options are not given), we can follow these general steps:

Step 1: Check the limit as \(x\) approaches \(1\) from the left (\(x

ightarrow1^{-}\))
For \(0\lt x\lt1\), \(f(x)=3^{x}\). We use the limit formula \(\lim_{x
ightarrow a}a^{x}=a^{a}\) (when \(a > 0\)).
\(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}3^{x}=3^{1} = 3\)

Step 2: Check the limit as \(x\) approaches \(1\) from the right (\(x

ightarrow1^{+}\))
For \(1\lt x\lt2\), \(f(x)=\frac{1}{2}x^{2}-x+\frac{7}{2}\). We substitute \(x = 1\) into this function (since it is a polynomial and continuous everywhere in its domain).
\(\lim_{x
ightarrow1^{+}}f(x)=\frac{1}{2}(1)^{2}-1+\frac{7}{2}=\frac{1}{2}-1 + \frac{7}{2}=\frac{1 - 2+7}{2}=\frac{6}{2}=3\)

Step 3: Check the value of the function at \(x = 1\) (but note that the function is not defined at \(x = 1\) in the given piece - wise definition, since the domains are \(0\lt x\lt1\) and \(1\lt x\lt2\))

If we want to check continuity at \(x = 1\), a function \(y = f(x)\) is continuous at \(x=a\) if \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)\). Since \(f(1)\) is not defined, but \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x) = 3\), the function has a removable discontinuity at \(x = 1\) (if we define \(f(1)=3\), the function becomes continuous at \(x = 1\)).

If we want to check differentiability at \(x = 1\), first, a function must be continuous at a point to be differentiable there. Since the function is not defined at \(x = 1\) (in the given domain), but the left - hand and right - hand limits exist and are equal, if we extend the function to be defined at \(x = 1\) as \(f(1)=3\), we can then check the derivatives.

The derivative of \(y = 3^{x}\) is \(y^\prime=3^{x}\ln(3)\) (using the formula \(\frac{d}{dx}(a^{x})=a^{x}\ln(a)\)). The left - hand derivative at \(x = 1\) is \(f^\prime(1^{-})=3^{1}\ln(3)=3\ln(3)\approx3\times1.0986 = 3.2958\)

The derivative of \(y=\frac{1}{2}x^{2}-x+\frac{7}{2}\) is \(y^\prime=x - 1\) (using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\)). The right - hand derivative at \(x = 1\) is \(f^\prime(1^{+})=1-1 = 0\)

Since \(f^\prime(1^{-})
eq f^\prime(1^{+})\) (even if we define \(f(1) = 3\) to make it continuous), the function is not differentiable at \(x = 1\) (if we consider the extended function with \(f(1)=3\)).

However, since the options are not provided, we can only give the general method of analyzing the piece - wise function at the point \(x = 1\) (the point where the two pieces of the function meet). If the question was about continuity, we would say that \(\lim_{x
ightarrow1}f(x)=3\) (since left - hand and right - hand limits are equal), and if the question was about differentiability, we would say that the function is not differentiable at \(x = 1\) (because the left - hand and right - hand derivatives are not equal).

If you can provide the options, we can give a more specific answer.

Answer:

Since the problem is about a piece - wise function and we need to analyze its properties (probably continuity or differentiability at \(x = 1\), but the options are not given), we can follow these general steps:

Step 1: Check the limit as \(x\) approaches \(1\) from the left (\(x

ightarrow1^{-}\))
For \(0\lt x\lt1\), \(f(x)=3^{x}\). We use the limit formula \(\lim_{x
ightarrow a}a^{x}=a^{a}\) (when \(a > 0\)).
\(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}3^{x}=3^{1} = 3\)

Step 2: Check the limit as \(x\) approaches \(1\) from the right (\(x

ightarrow1^{+}\))
For \(1\lt x\lt2\), \(f(x)=\frac{1}{2}x^{2}-x+\frac{7}{2}\). We substitute \(x = 1\) into this function (since it is a polynomial and continuous everywhere in its domain).
\(\lim_{x
ightarrow1^{+}}f(x)=\frac{1}{2}(1)^{2}-1+\frac{7}{2}=\frac{1}{2}-1 + \frac{7}{2}=\frac{1 - 2+7}{2}=\frac{6}{2}=3\)

Step 3: Check the value of the function at \(x = 1\) (but note that the function is not defined at \(x = 1\) in the given piece - wise definition, since the domains are \(0\lt x\lt1\) and \(1\lt x\lt2\))

If we want to check continuity at \(x = 1\), a function \(y = f(x)\) is continuous at \(x=a\) if \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)\). Since \(f(1)\) is not defined, but \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x) = 3\), the function has a removable discontinuity at \(x = 1\) (if we define \(f(1)=3\), the function becomes continuous at \(x = 1\)).

If we want to check differentiability at \(x = 1\), first, a function must be continuous at a point to be differentiable there. Since the function is not defined at \(x = 1\) (in the given domain), but the left - hand and right - hand limits exist and are equal, if we extend the function to be defined at \(x = 1\) as \(f(1)=3\), we can then check the derivatives.

The derivative of \(y = 3^{x}\) is \(y^\prime=3^{x}\ln(3)\) (using the formula \(\frac{d}{dx}(a^{x})=a^{x}\ln(a)\)). The left - hand derivative at \(x = 1\) is \(f^\prime(1^{-})=3^{1}\ln(3)=3\ln(3)\approx3\times1.0986 = 3.2958\)

The derivative of \(y=\frac{1}{2}x^{2}-x+\frac{7}{2}\) is \(y^\prime=x - 1\) (using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\)). The right - hand derivative at \(x = 1\) is \(f^\prime(1^{+})=1-1 = 0\)

Since \(f^\prime(1^{-})
eq f^\prime(1^{+})\) (even if we define \(f(1) = 3\) to make it continuous), the function is not differentiable at \(x = 1\) (if we consider the extended function with \(f(1)=3\)).

However, since the options are not provided, we can only give the general method of analyzing the piece - wise function at the point \(x = 1\) (the point where the two pieces of the function meet). If the question was about continuity, we would say that \(\lim_{x
ightarrow1}f(x)=3\) (since left - hand and right - hand limits are equal), and if the question was about differentiability, we would say that the function is not differentiable at \(x = 1\) (because the left - hand and right - hand derivatives are not equal).

If you can provide the options, we can give a more specific answer.