$$ f(x)=\frac{4 x^{3}}{3}+5 x^{2}-6 x \quad $$ on $$ [-4,1] $$

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1. Find the derivative of 
   $$
   f(x)=\frac{4x^3}{3}+5x^2-6x.
   $$
   We have
   $$
   f'(x)=\frac{d}{dx}\left(\frac{4x^3}{3}\right)+\frac{d}{dx}(5x^2)-\frac{d}{dx}(6x)
   =4x^2+10x-6.
   $$

2. Set the derivative equal to zero to find the critical points:
   $$
   4x^2+10x-6=0.
   $$
   Use the quadratic formula:
   $$
   x=\frac{-10\pm\sqrt{10^2-4\cdot4\cdot(-6)}}{2\cdot4}
   =\frac{-10\pm\sqrt{100+96}}{8}
   =\frac{-10\pm\sqrt{196}}{8}
   =\frac{-10\pm14}{8}.
   $$
   So,
   $$
   x=\frac{-10+14}{8}=\frac{4}{8}=\frac{1}{2}
   $$
   and
   $$
   x=\frac{-10-14}{8}=-\frac{24}{8}=-3.
   $$

3. Evaluate $$ f(x) $$ at the endpoints and at the critical points within the interval $$[-4,1]$$.

- At $$ x=-4 $$:
     $$
     f(-4)=\frac{4(-4)^3}{3}+5(-4)^2-6(-4)
     =\frac{4(-64)}{3}+5\cdot16+24
     =-\frac{256}{3}+80+24.
     $$
     Write $$80+24=104$$:
     $$
     f(-4)=-\frac{256}{3}+\frac{312}{3}=\frac{56}{3}.
     $$
     
   - At $$ x=-3 $$:
     $$
     f(-3)=\frac{4(-3)^3}{3}+5(-3)^2-6(-3)
     =\frac{4(-27)}{3}+5\cdot9+18
     =-36+45+18=27.
     $$
     
   - At $$ x=\frac{1}{2} $$:
     $$
     f\left(\frac{1}{2}\right)=\frac{4\left(\frac{1}{2}\right)^3}{3}+5\left(\frac{1}{2}\right)^2-6\left(\frac{1}{2}\right)
     =\frac{4\left(\frac{1}{8}\right)}{3}+5\left(\frac{1}{4}\right)-3
     =\frac{4}{24}+\frac{5}{4}-3.
     $$
     Simplify:
     $$
     \frac{4}{24}=\frac{1}{6}.
     $$
     Write all terms with a common denominator (12):
     $$
     \frac{1}{6}=\frac{2}{12},\quad\frac{5}{4}=\frac{15}{12},\quad 3=\frac{36}{12}.
     $$
     So,
     $$
     f\left(\frac{1}{2}\right)=\frac{2}{12}+\frac{15}{12}-\frac{36}{12}
     =\frac{2+15-36}{12}
     =\frac{-19}{12}.
     $$

- At $$ x=1 $$:
     $$
     f(1)=\frac{4(1)^3}{3}+5(1)^2-6(1)
     =\frac{4}{3}+5-6
     =\frac{4}{3}-1
     =\frac{4-3}{3}
     =\frac{1}{3}.
     $$

4. Compare the values:
   $$
   f(-4)=\frac{56}{3}\approx18.67,\quad f(-3)=27,\quad f\left(\frac{1}{2}\right)=-\frac{19}{12}\approx-1.58,\quad f(1)=\frac{1}{3}\approx0.33.
   $$
   The absolute maximum is $$ 27 $$ at $$ x=-3 $$, and the absolute minimum is $$ -\frac{19}{12} $$ at $$ x=\frac{1}{2} $$.
The function $$ f(x) = \frac{4x^3}{3} + 5x^2 - 6x $$ on the interval $$[-4,1]$$ has an absolute maximum of 27 at $$ x = -3 $$ and an absolute minimum of $$-\frac{19}{12}$$ at $$ x = \frac{1}{2} $$.

\( f(x)=\frac{4 x^{3}}{3}+5 x^{2}-6 x \quad \) on \( [-4,1] \)

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