QuestionFind $c$ for the function defined by $f(x) = x^2 - 4$ if $x \leq c$ and $f(x) = 6x - 13$ if $x > c$ to ensure continuity. $c =$

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is defined as a piecewise function with two parts $x^{}-4$ for $x \leq c$ and $6x-13$ for $x>c$ . . We need to find the value of $c$ such that the function is continuous everywhere.

STEP 2

A function is continuous at a point if the limit of the function as $x$ approaches that point from the left equals the limit of the function as $x$ approaches that point from the right, and both of these limits are equal to the function's value at that point.So, for the function to be continuous at $x=c$ , we have to ensure that $\lim{{x \to c^-}} f(x) = \lim{{x \to c^+}} f(x) = f(c)$

STEP 3

First, let's find the left-hand limit, which is the limit of the function as $x$ approaches $c$ from the left. This corresponds to the first part of the piecewise function, $x^{2}-$ .
$\lim{{x \to c^-}} f(x) = \lim{{x \to c^-}} (x^{2}-)$

STEP 4

Plug in the value $c$ into the expression to calculate the left-hand limit.
$\lim{{x \to c^-}} f(x) = c^{2}-4$

STEP 5

Next, let's find the right-hand limit, which is the limit of the function as $x$ approaches $c$ from the right. This corresponds to the second part of the piecewise function, $x-13$ .
$\lim{{x \to c^+}} f(x) = \lim{{x \to c^+}} (x-13)$

STEP 6

Plug in the value $c$ into the expression to calculate the right-hand limit.
$\lim{{x \to c^+}} f(x) =6c-13$

STEP 7

Now, we need to set the left-hand limit equal to the right-hand limit and solve for $c$ .
$c^{2}-4 =6c-13$

STEP 8

Rearrange the equation to form a quadratic equation.
$c^{2} -6c +4 +13 =0$

STEP 9

implify the equation.
$c^{2} -6c +17 =$

STEP 10

Now, solve the quadratic equation for $c$ . We can use the quadratic formula, which is given by $c = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}$ where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation.

STEP 11

Plug in the values for $a$ , $b$ , and $c$ into the quadratic formula.
$c = \frac{-(-6) \pm \sqrt{(-6)^{} -4**17}}{*}$

STEP 12

implify the equation.
$c = \frac{6 \pm \sqrt{36 -68}}{2}$

STEP 13

Further simplify the equation.
$c = \frac{6 \pm \sqrt{-32}}{2}$

STEP 14

Since the square root of a negative number is not a real number, there are no real solutions for $c$ . Therefore, the function $f(x)$ cannot be continuous everywhere for any real value of $c$ .
The solution is that there is no real value of $c$ for which the function $f(x)$ is continuous everywhere.

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