Question

Let f be the function given by f(x) = x+4(x−1)(x+3) on the closed interval [−5,5]. On which closed interval is the function f guaranteed by the Extreme Value Theorem to have an absolute maximum and an absolute minimum?

Answer 1

Answer:

Step-by-step explanation:

f(x)=x+4(x^2+2x-3)=4x^2+9x-12

f'(x)=8x+9

f'(x)=0,gives x=-9/8

f(-5)=-5+4(-5-1)(-5+3)=-5+4*-6*-2=43

f(-9/8)=-9/8+4(-9/8-1)(-9/8+3)

=-9/8+4*-17/8*15/8

=-9/8-255/16

=-273/16=-17 1/16

f(5)=4*5^2+9*5-12=100+45-12=133

absolute maximum=133

absolute minimum=-17 1/16

Answer 2

For given function, absolute maxima at x = 5 and absolute minima at $x=-\frac{9}{8}$

Absolute maximum value is 133 and absolute minimum value is $-\frac{273}{16}$

Given function is,

                 $f(x)=x+4(x-1)(x+3)\\\\f(x)=4x^{2} +9x-12$

To find critical points, Differentiate above function and equate with zero.

            $\frac{df}{dx} =8x+9=0\\\\x=-\frac{9}{8}$

We have critical points are, $x=5,x=-\frac{9}{8} ,x=-5$

                        $f(-5)=43\\\\f(-\frac{9}{8} )=-\frac{273}{16} \\\\f(5)=133$

From above, it is clear that $f(5)$ is absolute maximum value and $f(-\frac{9}{8} )$ is absolute minimum value.

Thus, For given function, absolute maxima at x = 5 and absolute minima at $x=-\frac{9}{8}$

Absolute maximum value is 133 and absolute minimum value is $-\frac{273}{16}$

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