Answer:
continuous and differentiatable
Step-by-step explanation:
We are given a piecewise function:
[tex]\displaystyle \large{f(x)=\left \{ {{4x-1 \ \ \ (x>1)} \atop {x^2+2x \ \ \ (x\leq 1)}} \right. }[/tex]
Determine whether if a function is continuous at x = 1 or not, depending on the following condition. A function that can be continuous as a specific point has to satisfy the followings:
- Both left and right limits are equal which leads to limit having a value and is finite.
- The specific point or f(a) can be found, evaluated and is finite.
- The limit as x tends to a equals to f(a).
With the three conditions above, we can conclude that if a function at x = a does satisfy them, it’s a continuous function at that point.
First, we have to find Limit - consider both sides of Limit which are left and right. Apply left limit with x^2+2x because the domain is less or equal to and apply right limit to 4x-1 because the domain is greater than 1.
Finding Limit at x = 1
[tex]\displaystyle \large{ \lim_{x \to 1^-} x^2+2x}=(1)^2+2(1) \to 1+2= \boxed{3}}[/tex]
So our left limit value is 3.
[tex]\displaystyle \large{ \lim_{x \to 1^+} 4x-1 = 4(1)-1 \to 4-1=\boxed{3}}[/tex]
The right limit value is 3 as well.
Since both limits are equal, therefore:
[tex]\displaystyle \large \boxed{ \lim_{x \to 1} f(x) = 3}[/tex]
Hence, it satisfies the first condition where the limit must exist and is finite.
Finding f(1)
Next is to see if f(1) can be evaluated or not. If it can be evaluated and is finite, it’ll satisfy the second condition but it must also have to equal the limit value as well to satisfy the third condition.
The domain of x^2+2x is perfect for substituting x = 1 in because the function has domain of x being equal or less than 1 but another function has domain greater than 1.
So if we substitute f(1) in.x^2+2x, we obtain 1+2(1) = 3.
Therefore, it satisfies the second and third condition.
Hence, the function is continuous at x = 1.
Next is to find whether if the function is differentiatable at x = 1 or not. All differentiatable functions are continuous but not all continuous functions are differentiatable.
The condition of differentiatable are:
- The specific point must exist and is finite.
- The function must be continuous.
- Limit of Derivative of both left and right are equal and thus making the limit exist.
However, the third condition is not necessary to use the limit which will consume your time a lot taking derivative limit. Instead, use the differentiation power rules and derive both sides then substitute x = 1 in to check if they are equal.
The first condition of derivative is already cleared, we know x = 1 can be evaluated as well as second, we just proved that a function was continuous. The left is to derive both functions then substitute x = 1 in.
Power Rules
[tex]\displaystyle \large{y=ax^n \to y\prime = nax^{n-1}}[/tex]
These functions are polynomial so we just use power rules, not really complicated.
From the piecewise function, derive both sides and take left derivative and right derivative.
Left Derivative
[tex]\displaystyle \large{y=x^2+2x \to y\prime = 2x+2}[/tex]
Then substitute x = 1 as we obtain 2(1) + 2 = 4
Right Derivative
[tex]\displaystyle \large{y = 4x-1 \to y\prime = 4}[/tex]
Since there is no x-term to substitute, the right derivative value is 4.
Since left derivative and right derivative are equal, the derivative limit exists and equal 4.
Hence, all conditions are satisfied and therefore, the function is differentiatable at x = 1.
So, the function is continuous and differentiatable at x = 1.
