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Math  /  Calculus

QuestionConsider the function f(x)=x4−98x2+11f(x)=x^{4}-98 x^{2}+11 defined on the closed interval [−6,15][-6,15]. This function has an absolute minimum value equal to □\square and an absolute maximum value equal to □\square

Studdy Solution

STEP 1

1. The function f(x)=x4−98x2+11 f(x) = x^4 - 98x^2 + 11 is a polynomial, which is continuous and differentiable everywhere.
2. To find the absolute extrema (minimum and maximum) on a closed interval, we need to evaluate the function at critical points and endpoints of the interval.
3. Critical points occur where the derivative is zero or undefined.

STEP 2

1. Find the derivative of the function.
2. Determine the critical points within the interval.
3. Evaluate the function at the critical points and endpoints.
4. Compare the values to find the absolute minimum and maximum.

STEP 3

Find the derivative of the function f(x) f(x) .
f(x)=x4−98x2+11 f(x) = x^4 - 98x^2 + 11
The derivative f′(x) f'(x) is obtained by differentiating each term:
f′(x)=4x3−196x f'(x) = 4x^3 - 196x

STEP 4

Determine the critical points by setting the derivative equal to zero and solving for x x .
4x3−196x=0 4x^3 - 196x = 0
Factor out the common term:
4x(x2−49)=0 4x(x^2 - 49) = 0
This gives us:
4x=0orx2−49=0 4x = 0 \quad \text{or} \quad x^2 - 49 = 0
Solving these equations:
x=0 x = 0
x2=49 x^2 = 49
x=±7 x = \pm 7
The critical points are x=0,7,−7 x = 0, 7, -7 .

STEP 5

Evaluate the function f(x) f(x) at the critical points and endpoints of the interval [−6,15][-6, 15].
f(−6)=(−6)4−98(−6)2+11 f(-6) = (-6)^4 - 98(-6)^2 + 11 f(0)=04−98(0)2+11 f(0) = 0^4 - 98(0)^2 + 11 f(7)=74−98(7)2+11 f(7) = 7^4 - 98(7)^2 + 11 f(−7)=(−7)4−98(−7)2+11 f(-7) = (-7)^4 - 98(-7)^2 + 11 f(15)=154−98(15)2+11 f(15) = 15^4 - 98(15)^2 + 11
Calculate each value:
f(−6)=1296−3528+11=−2221 f(-6) = 1296 - 3528 + 11 = -2221 f(0)=11 f(0) = 11 f(7)=2401−4802+11=−2390 f(7) = 2401 - 4802 + 11 = -2390 f(−7)=2401−4802+11=−2390 f(-7) = 2401 - 4802 + 11 = -2390 f(15)=50625−22050+11=28586 f(15) = 50625 - 22050 + 11 = 28586

STEP 6

Compare the values obtained to determine the absolute minimum and maximum.
The values are: - f(−6)=−2221 f(-6) = -2221 - f(0)=11 f(0) = 11 - f(7)=−2390 f(7) = -2390 - f(−7)=−2390 f(-7) = -2390 - f(15)=28586 f(15) = 28586
The absolute minimum value is −2390 -2390 at x=7 x = 7 and x=−7 x = -7 .
The absolute maximum value is 28586 28586 at x=15 x = 15 .
The absolute minimum value is −2390\boxed{-2390} and the absolute maximum value is 28586\boxed{28586}.

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