Question

estimate the y - coordinate at which the relative maxima and relative minima occur for the function. 20. f(x)=-x^4 + 2x^2+3 21. f(x)=0.75x^4 + x^3 - 3x^2+4

Explanation:

Step1: Find the derivative of the function $f(x)=-x^{4}+2x^{2}+3$.

Using the power - rule $(x^n)^\prime=nx^{n - 1}$, we have $f^\prime(x)=-4x^{3}+4x$.

Step2: Set the derivative equal to zero to find the critical points.

$-4x^{3}+4x = 0$. Factor out $-4x$: $-4x(x^{2}-1)=0$. Then $-4x(x - 1)(x + 1)=0$. So the critical points are $x = 0,x = 1,x=-1$.

Step3: Use the second - derivative test to determine if the critical points are relative maxima or minima.

Find the second - derivative $f^{\prime\prime}(x)=-12x^{2}+4$.
When $x = 0$, $f^{\prime\prime}(0)=4>0$, so $f(x)$ has a relative minimum at $x = 0$. And $f(0)=-0^{4}+2\times0^{2}+3 = 3$.
When $x = 1$, $f^{\prime\prime}(1)=-12\times1^{2}+4=-8<0$, so $f(x)$ has a relative maximum at $x = 1$. And $f(1)=-1^{4}+2\times1^{2}+3=-1 + 2+3=4$.
When $x=-1$, $f^{\prime\prime}(-1)=-12\times(-1)^{2}+4=-8<0$, so $f(x)$ has a relative maximum at $x=-1$. And $f(-1)=-(-1)^{4}+2\times(-1)^{2}+3=-1 + 2+3=4$.

For the function $f(x)=0.75x^{4}+x^{3}-3x^{2}+4$:

Step1: Find the first - derivative.

$f^\prime(x)=3x^{3}+3x^{2}-6x=3x(x^{2}+x - 2)=3x(x + 2)(x - 1)$.

Step2: Find the critical points.

Set $f^\prime(x)=0$. Then $3x(x + 2)(x - 1)=0$. So the critical points are $x = 0,x=-2,x = 1$.

Step3: Find the second - derivative.

$f^{\prime\prime}(x)=9x^{2}+6x-6$.
When $x = 0$, $f^{\prime\prime}(0)=-6<0$, so $f(x)$ has a relative maximum at $x = 0$. And $f(0)=0.75\times0^{4}+0^{3}-3\times0^{2}+4 = 4$.
When $x=-2$, $f^{\prime\prime}(-2)=9\times(-2)^{2}+6\times(-2)-6=36-12 - 6=18>0$, so $f(x)$ has a relative minimum at $x=-2$. And $f(-2)=0.75\times(-2)^{4}+(-2)^{3}-3\times(-2)^{2}+4=0.75\times16-8 - 12 + 4=12-8 - 12 + 4=-4$.
When $x = 1$, $f^{\prime\prime}(1)=9\times1^{2}+6\times1-6=9 + 6-6=9>0$, so $f(x)$ has a relative minimum at $x = 1$. And $f(1)=0.75\times1^{4}+1^{3}-3\times1^{2}+4=0.75+1-3 + 4=2.75$.

Answer:

For $f(x)=-x^{4}+2x^{2}+3$: Relative minimum at $(0,3)$, relative maxima at $(1,4)$ and $(-1,4)$.
For $f(x)=0.75x^{4}+x^{3}-3x^{2}+4$: Relative maximum at $(0,4)$, relative minima at $(-2,-4)$ and $(1,2.75)$.