The function $$ f(x)=-2 x^{3}+33 x^{2}-180 x+3 $$ has one local minimum and one local maximum. This function has a local minimum at $$ x=\square $$ with value and a local maximum at $$ x= $$ with value Question Help: $$ \square $$ Video

Question

UpStudy AI · Accepted Answer

To find the local minimum and maximum of the function $$ f(x) = -2x^3 + 33x^2 - 180x + 3 $$, we need to find the critical points of the function. Critical points are the points where the derivative of the function is equal to zero or undefined. Let's find the derivative of the function $$ f(x) $$ with respect to $$ x $$ and then set it equal to zero to find the critical points. The derivative of $$ f(x) $$ with respect to $$ x $$ is given by: $$ f'(x) = \frac{d}{dx}(-2x^3 + 33x^2 - 180x + 3) $$ Using the power rule for differentiation, we get: $$ f'(x) = -6x^2 + 66x - 180 $$ Now, we set $$ f'(x) $$ equal to zero to find the critical points: $$ -6x^2 + 66x - 180 = 0 $$ Let's solve this quadratic equation to find the critical points. Solve the quadratic equation by following steps: - step0: Solve by factoring: $$-6x^{2}+66x-180=0$$ - step1: Factor the expression: $$-6\left(x-6\right)\left(x-5\right)=0$$ - step2: Divide the terms: $$\left(x-6\right)\left(x-5\right)=0$$ - step3: Separate into possible cases: $$\begin{align}&x-6=0\&x-5=0\end{align}$$ - step4: Solve the equation: $$\begin{align}&x=6\&x=5\end{align}$$ - step5: Rewrite: $$x_{1}=5,x_{2}=6$$ The critical points of the function $$ f(x) = -2x^3 + 33x^2 - 180x + 3 $$ are $$ x = 5 $$ and $$ x = 6 $$. To determine whether these critical points correspond to local minima or maxima, we need to examine the second derivative of the function. The second derivative of $$ f(x) $$ with respect to $$ x $$ is given by: $$ f''(x) = \frac{d}{dx}(-6x^2 + 66x - 180) $$ Using the power rule for differentiation, we get: $$ f''(x) = -12x + 66 $$ Now, we evaluate the second derivative at the critical points to determine the nature of the critical points. At $$ x = 5 $$: $$ f''(5) = -12(5) + 66 = -60 + 66 = 6 > 0 $$ Since $$ f''(5) > 0 $$, the critical point $$ x = 5 $$ corresponds to a local minimum. At $$ x = 6 $$: $$ f''(6) = -12(6) + 66 = -72 + 66 = -6 < 0 $$ Since $$ f''(6) < 0 $$, the critical point $$ x = 6 $$ corresponds to a local maximum. Therefore, the local minimum of the function $$ f(x) = -2x^3 + 33x^2 - 180x + 3 $$ is at $$ x = 5 $$ with a value of $$ f(5) $$, and the local maximum is at $$ x = 6 $$ with a value of $$ f(6) $$. The function $$ f(x) = -2x^3 + 33x^2 - 180x + 3 $$ has a local minimum at $$ x = 5 $$ and a local maximum at $$ x = 6 $$.

The function has
one local minimum and one local maximum.
This function has a local minimum at
with value
and a local maximum at
with value
Question Help: Video

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Calculus Questions

The function has one local minimum and one local maximum. This function has a local minimum at with value and a local maximum at with value Question Help: Video