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Math  /  Calculus

Question3. Find the a bsolute max and min f(x)=x32−3x,0≤x≤4f(x)=x^{\frac{3}{2}}-3 \sqrt{x}, 0 \leq x \leq 4 f(x)=x32−3x12f′(x)=32x12−32x−12\begin{array}{l} f(x)=x^{\frac{3}{2}}-\frac{3 x^{\frac{1}{2}}}{} \\ f^{\prime}(x)=\frac{3}{2} x^{\frac{1}{2}}-\frac{3}{2} x^{-\frac{1}{2}} \end{array}

Studdy Solution

STEP 1

1. The function f(x)=x32−3x f(x) = x^{\frac{3}{2}} - 3\sqrt{x} is continuous on the interval [0,4][0, 4].
2. The absolute maximum and minimum values of a continuous function on a closed interval occur either at critical points or at the endpoints of the interval.
3. Critical points occur where the derivative is zero or undefined.

STEP 2

1. Verify the given derivative.
2. Find critical points by setting the derivative equal to zero.
3. Evaluate the function at critical points and endpoints.
4. Determine the absolute maximum and minimum values.

STEP 3

Verify the given derivative of the function:
The function is f(x)=x32−3x12 f(x) = x^{\frac{3}{2}} - 3x^{\frac{1}{2}} .
The derivative is given as:
f′(x)=32x12−32x−12 f'(x) = \frac{3}{2} x^{\frac{1}{2}} - \frac{3}{2} x^{-\frac{1}{2}}
This derivative is correct as it follows the power rule for differentiation.

STEP 4

Find critical points by setting the derivative f′(x) f'(x) equal to zero:
32x12−32x−12=0 \frac{3}{2} x^{\frac{1}{2}} - \frac{3}{2} x^{-\frac{1}{2}} = 0
Factor out 32\frac{3}{2}:
32(x12−x−12)=0 \frac{3}{2} (x^{\frac{1}{2}} - x^{-\frac{1}{2}}) = 0
Simplify the expression inside the parentheses:
x12−x−12=0 x^{\frac{1}{2}} - x^{-\frac{1}{2}} = 0
Multiply through by x12 x^{\frac{1}{2}} to eliminate the negative exponent:
x−1=0 x - 1 = 0
Solve for x x :
x=1 x = 1

STEP 5

Evaluate the function f(x) f(x) at the critical point and the endpoints of the interval [0,4][0, 4]:
1. At x=0 x = 0 :
f(0)=032−30=0 f(0) = 0^{\frac{3}{2}} - 3\sqrt{0} = 0
2. At x=1 x = 1 :
f(1)=132−31=1−3=−2 f(1) = 1^{\frac{3}{2}} - 3\sqrt{1} = 1 - 3 = -2
3. At x=4 x = 4 :
f(4)=432−34=8−6=2 f(4) = 4^{\frac{3}{2}} - 3\sqrt{4} = 8 - 6 = 2

STEP 6

Determine the absolute maximum and minimum values from the evaluated points:
- The absolute minimum value is −2-2 at x=1 x = 1 . - The absolute maximum value is 22 at x=4 x = 4 .
The absolute maximum value is 2 \boxed{2} and the absolute minimum value is −2 \boxed{-2} .

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