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(A) Given $f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11$.Differentiating with respect to $x$:$f'(x) = \frac{3}{10}(4x^3) - \fra

Vedclass · Accepted Answer

Increasing: $(-2, 1) \cup (3, \infty)$,Decreasing: $(-\infty, -2) \cup (1, 3)$

Vedclass · Answer

(A) Given $f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11$.Differentiating with respect to $x$:$f'(x) = \frac{3}{10}(4x^3) - \frac{4}{5}(3x^2) - 3(2x) + \frac{36}{5}$$f'(x) = \frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5}$$f'(x) = \frac{6}{5}(x^3 - 2x^2 - 5x + 6)$Factoring the cubic polynomial:$f'(x) = \frac{6}{5}(x - 1)(x + 2)(x - 3)$Setting $f'(x) = 0$,we get critical points $x = 1, x = -2, x = 3$. These points divide the real line into intervals $(-\infty, -2), (-2, 1), (1, 3), (3, \infty)$.Testing the sign of $f'(x)$ in each interval:$1.$ For $(-\infty, -2)$,choose $x = -3$: $f'(-3) = \frac{6}{5}(-4)(-1)(-6) $2.$ For $(-2, 1)$,choose $x = 0$: $f'(0) = \frac{6}{5}(-1)(2)(-3) > 0$. Thus,$f$ is increasing.$3.$ For $(1, 3)$,choose $x = 2$: $f'(2) = \frac{6}{5}(1)(4)(-1) $4.$ For $(3, \infty)$,choose $x = 4$: $f'(4) = \frac{6}{5}(3)(6)(1) > 0$. Thus,$f$ is increasing.Conclusion:$(a)$ The function is increasing in $(-2, 1) \cup (3, \infty)$.$(b)$ The function is decreasing in $(-\infty, -2) \cup (1, 3)$.

Find the intervals in which the function given by $f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11$ is $(a)$ increasing $(b)$ decreasing.

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