Find the absolute maximum and absolute minimum values of $$ f $$ on the given interval. $$ f(x)=2 x^{3}-6 x^{2}-48 x+5,[-3,5] $$ absolute minimum

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It seems there was an issue with the calculation. Let's try to find the critical points manually by taking the derivative of the function and setting it equal to zero.

The derivative of $$ f(x) = 2x^{3} - 6x^{2} - 48x + 5 $$ is $$ f'(x) = 6x^{2} - 12x - 48 $$.

Setting $$ f'(x) = 0 $$, we have:
$$ 6x^{2} - 12x - 48 = 0 $$

Solving this quadratic equation, we find the critical points. Let's solve it manually.
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
  $$6x^{2}-12x-48=0$$
- step1: Factor the expression:
  $$6\left(x-4\right)\left(x+2\right)=0$$
- step2: Divide the terms:
  $$\left(x-4\right)\left(x+2\right)=0$$
- step3: Separate into possible cases:
  $$\begin{align}&x-4=0\&x+2=0\end{align}$$
- step4: Solve the equation:
  $$\begin{align}&x=4\&x=-2\end{align}$$
- step5: Rewrite:
  $$x_{1}=-2,x_{2}=4$$
The critical points of the function are $$ x = -2 $$ and $$ x = 4 $$.

Next, we need to evaluate the function at the critical points and the endpoints of the interval $$[-3, 5]$$ to find the absolute maximum and absolute minimum values.
Evaluate the expression by following steps:
- step0: Evaluate:
  $$2x^{3}-6x^{2}-48x+5$$
- step1: Substitute:
  $$2\left(-2\right)^{3}-6\left(-2\right)^{2}-48\left(-2\right)+5$$
- step2: Multiply the terms:
  $$-2^{4}-6\left(-2\right)^{2}-48\left(-2\right)+5$$
- step3: Multiply the terms:
  $$-2^{4}-24-48\left(-2\right)+5$$
- step4: Multiply the numbers:
  $$-2^{4}-24+96+5$$
- step5: Evaluate the power:
  $$-16-24+96+5$$
- step6: Calculate:
  $$61$$
Substitute $$ x=4 $$ into the expression $$ 2x^{3}-6x^{2}-48x+5 $$.
Evaluate the expression by following steps:
- step0: Evaluate:
  $$2x^{3}-6x^{2}-48x+5$$
- step1: Substitute:
  $$2\times 4^{3}-6\times 4^{2}-48\times 4+5$$
- step2: Multiply the terms:
  $$128-6\times 4^{2}-48\times 4+5$$
- step3: Multiply the terms:
  $$128-96-48\times 4+5$$
- step4: Multiply the numbers:
  $$128-96-192+5$$
- step5: Calculate:
  $$-155$$
Substitute $$ x=-3 $$ into the expression $$ 2x^{3}-6x^{2}-48x+5 $$.
Evaluate the expression by following steps:
- step0: Evaluate:
  $$2x^{3}-6x^{2}-48x+5$$
- step1: Substitute:
  $$2\left(-3\right)^{3}-6\left(-3\right)^{2}-48\left(-3\right)+5$$
- step2: Multiply the terms:
  $$-54-6\left(-3\right)^{2}-48\left(-3\right)+5$$
- step3: Multiply the terms:
  $$-54-54-48\left(-3\right)+5$$
- step4: Multiply the numbers:
  $$-54-54+144+5$$
- step5: Calculate:
  $$41$$
Substitute $$ x=5 $$ into the expression $$ 2x^{3}-6x^{2}-48x+5 $$.
Evaluate the expression by following steps:
- step0: Evaluate:
  $$2x^{3}-6x^{2}-48x+5$$
- step1: Substitute:
  $$2\times 5^{3}-6\times 5^{2}-48\times 5+5$$
- step2: Multiply the terms:
  $$250-6\times 5^{2}-48\times 5+5$$
- step3: Multiply the terms:
  $$250-150-48\times 5+5$$
- step4: Multiply the numbers:
  $$250-150-240+5$$
- step5: Calculate:
  $$-135$$
The values of the function at the critical points and the endpoints are:
- At $$ x = -2 $$, $$ f(-2) = 61 $$
- At $$ x = 4 $$, $$ f(4) = -155 $$
- At $$ x = -3 $$, $$ f(-3) = 41 $$
- At $$ x = 5 $$, $$ f(5) = -135 $$

Comparing these values, we can determine the absolute maximum and absolute minimum values of the function on the interval $$[-3, 5]$$.
The absolute maximum value of $$ f(x) $$ on $$[-3, 5]$$ is 61 at $$ x = -2 $$, and the absolute minimum value is -155 at $$ x = 4 $$.

Find the absolute maximum and absolute minimum values of \( f \) on the given interval. \[ f(x)=2 x^{3}-6 x^{2}-48 x+5,[-3,5] \] absolute minimum

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