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Determine whether the function 
f(x) is continuous at 
x=3.

f(x)={[4-3x^(2)",",x > 3],[-13-3x",",x <= 3]:}

f(x) is discontinuous at 
x=3

f(x) is continuous at 
x=3

Determine whether the function f(x) f(x) is continuous at x=3 x=3 .\newlinef(x)={43x2,amp;xgt;3133x,amp;x3 f(x)=\left\{\begin{array}{ll} 4-3 x^{2}, &amp; x&gt;3 \\ -13-3 x, &amp; x \leq 3 \end{array}\right. \newlinef(x) f(x) is discontinuous at x=3 x=3 \newlinef(x) f(x) is continuous at x=3 x=3

Full solution

Q. Determine whether the function f(x) f(x) is continuous at x=3 x=3 .\newlinef(x)={43x2,x>3133x,x3 f(x)=\left\{\begin{array}{ll} 4-3 x^{2}, & x>3 \\ -13-3 x, & x \leq 3 \end{array}\right. \newlinef(x) f(x) is discontinuous at x=3 x=3 \newlinef(x) f(x) is continuous at x=3 x=3
  1. Check Function Definition: To determine if the function f(x)f(x) is continuous at x=3x=3, we need to check three conditions:\newline11. The function is defined at x=3x=3.\newline22. The limit of f(x)f(x) as xx approaches 33 exists.\newline33. The limit of f(x)f(x) as xx approaches 33 is equal to the function value at x=3x=3.
  2. Find Left Limit: First, let's check if the function is defined at x=3x=3. We have two expressions for f(x)f(x), one for x > 3 and one for x3x \leq 3. Since we are interested in x=3x=3, we look at the expression for x3x \leq 3, which is f(x)=133xf(x) = -13 - 3x.f(3)=133(3)=139=22f(3) = -13 - 3(3) = -13 - 9 = -22.The function is defined at x=3x=3, and f(3)=22f(3) = -22.
  3. Find Right Limit: Next, we need to find the limit of f(x)f(x) as xx approaches 33 from the left (x3x \to 3^-). We use the expression for x3x \leq 3, which is f(x)=133xf(x) = -13 - 3x.limx3f(x)=limx3(133x)=133(3)=139=22\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (-13 - 3x) = -13 - 3(3) = -13 - 9 = -22.
  4. Determine Continuity: Now, we need to find the limit of f(x)f(x) as xx approaches 33 from the right (x3+x \to 3^+). We use the expression for x > 3, which is f(x)=43x2f(x) = 4 - 3x^2.limx3+f(x)=limx3+(43x2)=43(3)2=43(9)=427=23\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (4 - 3x^2) = 4 - 3(3)^2 = 4 - 3(9) = 4 - 27 = -23.
  5. Determine Continuity: Now, we need to find the limit of f(x)f(x) as xx approaches 33 from the right (x3+x \to 3^+). We use the expression for x > 3, which is f(x)=43x2f(x) = 4 - 3x^2.limx3+f(x)=limx3+(43x2)=43(3)2=43(9)=427=23\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (4 - 3x^2) = 4 - 3(3)^2 = 4 - 3(9) = 4 - 27 = -23.We have found that the limit from the left is 22-22 and the limit from the right is 23-23. Since these two limits are not equal, the limit of f(x)f(x) as xx approaches 33 does not exist. Therefore, the function f(x)f(x) is not continuous at xx33.

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