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Find all critical points of the function \newlinef(x)=9x+8x+1f(x)=9x+|8x+1|\newline(Use symbolic notation and fractions where needed. Give your answer in the form of a comma separated list. If the function does not have any critical points, enter DNE.)\newlinecritical points:

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Q. Find all critical points of the function \newlinef(x)=9x+8x+1f(x)=9x+|8x+1|\newline(Use symbolic notation and fractions where needed. Give your answer in the form of a comma separated list. If the function does not have any critical points, enter DNE.)\newlinecritical points:
  1. Define critical points: Understand the definition of critical points. Critical points occur where the derivative of the function is zero or undefined. We need to find the derivative of f(x)=9x+8x+1f(x) = 9x + |8x + 1| and solve for xx where the derivative is zero or does not exist.
  2. Calculate derivative: Calculate the derivative of f(x)f(x). The absolute value function 8x+1|8x + 1| is piecewise, so we need to consider two cases: when 8x+108x + 1 \geq 0 and when 8x + 1 < 0.\newlineCase 11: When 8x+108x + 1 \geq 0, 8x+1=8x+1|8x + 1| = 8x + 1, and the derivative f(x)=ddx(9x+8x+1)=17f'(x) = \frac{d}{dx} (9x + 8x + 1) = 17.\newlineCase 22: When 8x + 1 < 0, 8x+1=(8x+1)|8x + 1| = -(8x + 1), and the derivative f(x)=ddx(9x8x1)=1f'(x) = \frac{d}{dx} (9x - 8x - 1) = 1.
  3. Find derivative roots: Solve for xx where the derivative is zero. In Case 11, the derivative f(x)=17f'(x) = 17 is never zero. In Case 22, the derivative f(x)=1f'(x) = 1 is also never zero. Therefore, there are no critical points where the derivative is zero.
  4. Identify undefined points: Determine if there are any points where the derivative is undefined. The derivative is undefined at the point where the piecewise function changes from one case to the other, which is at 8x+1=08x + 1 = 0. Solve for xx: 8x+1=08x + 1 = 0, x=18.x = -\frac{1}{8}.
  5. Verify critical point: Verify that x=18x = -\frac{1}{8} is a critical point. Since the derivative changes from 1717 to 11 at x=18x = -\frac{1}{8}, this is indeed a critical point because the derivative does not exist at this point (the function has a corner or cusp).

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