Find a value for a so that the function f(x)={x2−1,x&lt;32axx≥3

f(x)={x2−1,x&lt;32axx≥3 Since the given function is apolynomial function for all x&lt;3 and x&gt;3, therefore it is continuous for all x&lt;3 and x&gt;3. Now we check the continuity at x=3. limx→3−f(x)=limx→3−(x2−1) =32−1 =8 and limx→3−=limx→3+f(x) ⇒8=6a ⇒a=43

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Jason Farmer

Answered question

2021-10-16

Find a value for a so that the function
$f (x) = {\begin{cases} x^{2} - 1, & x < 3 \\ 2 a x & x \geq 3 \end{cases}$

Answer & Explanation

Latisha Oneil

Skilled2021-10-17Added 100 answers

$f (x) = {\begin{cases} x^{2} - 1, & x < 3 \\ 2 a x & x \geq 3 \end{cases}$
Since the given function is apolynomial function for all x<3 and x>3, therefore it is continuous for all x<3 and x>3.
Now we check the continuity at x=3.
$lim_{x \to 3^{-}} f (x) = lim_{x \to 3^{-}} (x^{2} - 1)$
$= 3^{2} - 1$
$= 8$
and
$lim_{x \to 3^{-}} = lim_{x \to 3^{+}} f (x)$
$\Rightarrow 8 = 6 a$
$\Rightarrow a = \frac{4}{3}$

Jeffrey Jordon

Expert2022-08-24Added 2605 answers

Answer is given below (on video)

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