Math  /  Calculus

QuestionFind the intervals on which f(x)f(x) is increasing, the intervals on which f(x)f(x) is decreasing, and the local extrema. f(x)=x3+2x+1f(x)=x^{3}+2 x+1
Find f(x)f^{\prime}(x). f(x)=x3+2x+1f(x)=\begin{array}{l} f(x)=x^{3}+2 x+1 \\ f^{\prime}(x)=\square \end{array}

Studdy Solution

STEP 1

1. The function given is f(x)=x3+2x+1 f(x) = x^3 + 2x + 1 .
2. To find intervals of increase or decrease, we need the first derivative f(x) f'(x) .
3. Critical points occur where f(x)=0 f'(x) = 0 or f(x) f'(x) is undefined.
4. The sign of f(x) f'(x) determines whether f(x) f(x) is increasing or decreasing.

STEP 2

1. Differentiate f(x) f(x) to find f(x) f'(x) .
2. Solve f(x)=0 f'(x) = 0 to find critical points.
3. Determine the sign of f(x) f'(x) around the critical points.
4. Identify intervals of increase and decrease.
5. Determine local extrema based on sign changes in f(x) f'(x) .

STEP 3

Differentiate f(x)=x3+2x+1 f(x) = x^3 + 2x + 1 to find f(x) f'(x) .
f(x)=ddx(x3)+ddx(2x)+ddx(1) f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(2x) + \frac{d}{dx}(1)
f(x)=3x2+2 f'(x) = 3x^2 + 2

STEP 4

Solve f(x)=0 f'(x) = 0 to find critical points.
3x2+2=0 3x^2 + 2 = 0
3x2=2 3x^2 = -2
Since 3x2=2 3x^2 = -2 has no real solutions (as x2 x^2 cannot be negative), there are no critical points.

STEP 5

Determine the sign of f(x)=3x2+2 f'(x) = 3x^2 + 2 .
Since 3x2+2 3x^2 + 2 is always positive (as 3x20 3x^2 \geq 0 and adding 2 keeps it positive), f(x)>0 f'(x) > 0 for all x x .

STEP 6

Identify intervals of increase and decrease.
Since f(x)>0 f'(x) > 0 for all x x , f(x) f(x) is increasing on the interval (,)(-\infty, \infty).

STEP 7

Determine local extrema.
Since f(x) f'(x) does not change sign, there are no local extrema.
The function f(x)=x3+2x+1 f(x) = x^3 + 2x + 1 is increasing on the interval (,)(-\infty, \infty) and has no local extrema.

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