Question

the function f(x) = 2x^3 - 30x^2 + 126x + 3 has one local minimum and one local maximum. this function has a local minimum at x equals with value and a local maximum at x equals with value question help: post to forum submit question jump to answer

Explanation:

Step1: Find the derivative

Differentiate $f(x)=2x^{3}-30x^{2}+126x + 3$ using the power - rule. The derivative $f^\prime(x)=6x^{2}-60x + 126$.

Step2: Set the derivative equal to zero

Set $f^\prime(x)=0$, so $6x^{2}-60x + 126 = 0$. Divide through by 6 to get $x^{2}-10x + 21=0$.

Step3: Solve the quadratic equation

Factor $x^{2}-10x + 21=(x - 3)(x - 7)=0$. The solutions are $x = 3$ and $x = 7$.

Step4: Use the second - derivative test

Find the second - derivative $f^{\prime\prime}(x)=12x-60$.
When $x = 3$, $f^{\prime\prime}(3)=12\times3-60=-24<0$, so $x = 3$ is a local maximum.
When $x = 7$, $f^{\prime\prime}(7)=12\times7-60 = 24>0$, so $x = 7$ is a local minimum.

Step5: Find the function values

For the local minimum at $x = 7$, $f(7)=2\times7^{3}-30\times7^{2}+126\times7 + 3=2\times343-30\times49+882 + 3=686-1470+882 + 3=-1$.
For the local maximum at $x = 3$, $f(3)=2\times3^{3}-30\times3^{2}+126\times3 + 3=2\times27-30\times9+378 + 3=54-270+378 + 3=165$.

Answer:

This function has a local minimum at $x$ equals $7$
with value $-1$
and a local maximum at $x$ equals $3$
with value $165$