Let's find the first derivative to locate the relative maxima and minima.

$\displaystyle y=\frac{3}{4}x^4+3x^3-15x^2+8$

$\displaystyle y'=4*\frac{3}{4}x^3+9x^2-30x+0$

$\displaystyle y'=3x^3+9x^2-30x$

Now we set it equal to zero to find the x values of these critical points.

$\displaystyle 0=3x^3+9x^2-30x$

$\displaystyle 0=3x(x^2+3x-10)$

$\displaystyle 0=3x(x-2)(x+5)$

So the equation is 0 where x is -2, 0, or 5. Now let's find the second derivative so that we know which of these locations are maxima and which are minima.

$\displaystyle y'=3x^3+9x^2-30x$

$\displaystyle y''=9x^2+18x-30$

$\displaystyle y''(-5)=9(-5)^2+18(-5)-30=105>0$

So the function has a relative maximum at x=-5.

$\displaystyle y''(0)=9(0)^2+18(0)-30=-30< 0$

So the function has a relative minimum at x=0.

$\displaystyle y''(2)=9(2)^2+18(2)-30=42>0$

So the function has a relative maximum at x=2.

Thus there is only one relative minimum in this function, and it occurs at x=0. We need to plug this into the original function to find the y-coordinate of the point.

$\displaystyle y=\frac{3}{4}x^4+3x^3-15x^2+8$

$\displaystyle y=\frac{3}{4}(0)^4+3(0)^3-15(0)^2+8$

$\displaystyle y=0+0-0+8$

$\displaystyle y=8$

So our point is (0,8).

We need to find out where the first derivative is equal to zero to find the locations of the possible maxima and minima.

$\displaystyle f(x)=\frac{1}{2}x^4-x^2+3$

$\displaystyle f'(x)=\frac{4}{2}x^3-2x+0$

$\displaystyle f'(x)=2x^3-2x$

$\displaystyle f'(x)=2x(x^2-1)$

$\displaystyle f'(x)=2x(x+1)(x-1)$

$\displaystyle 0=2x(x+1)(x-1)$

$\displaystyle x=0,1,-1$ or $\displaystyle x=0,\pm1$.

We see that the function is a quadratic, so its graph will be a parabola. Thus, we know it will have only one relative max or min. Since the leading coefficient of $\displaystyle y=6x^2-2x+9$ is a positive 6, we also know that the parabola opens upward. Hence, the relative extremum is a minimum.

To take the derivative of a function, we'll need to apply the power rule to a term with a coefficient $\displaystyle A$ and an exponent $\displaystyle n$:

$\displaystyle \frac{d}{dx}(Ax^{n})=nAx^{n-1}$ 

Applying this rule to each term in the function, we start by taking the first derivative:

$\displaystyle f(x)=7x^{3}$

$\displaystyle f'(x)=3(7x^{3-1})$

$\displaystyle f'(x)=21x^{2}$

Taking the second derivative:

$\displaystyle f''(x)=2(21x^{2-1})$

$\displaystyle f''(x)=42x$

To take the derivative of a function, we'll need to apply the power rule to a term with a coefficient $\displaystyle A$ and an exponent $\displaystyle n$:

$\displaystyle \frac{d}{dx}(Ax^{n})=nAx^{n-1}$ 

Applying this rule to each term in the function, we start by taking the first derivative:

$\displaystyle f(x)=2x^{4}-12x^{3}+x^{2}-8x+5$

$\displaystyle f'(x)=4(2x^{4-1})-3(12x^{3-1})+2(x^{2-1})-1(8x^{1-1})+0(5)$

$\displaystyle f'(x)=8x^{3}-36x^{2}+2x-8$

Finally, we take the second derivative:

$\displaystyle f'(x)=8x^{3}-36x^{2}+2x-8x$

$\displaystyle f''(x)=3(8x^{3-1})-2(36x^{2-1})+1(2x^{1-1})-0(8)$

$\displaystyle f''(x)=24x^{2}-72x+2$

To take the derivative of a function, we'll need to apply the power rule to a term with a coefficient $\displaystyle A$ and an exponent $\displaystyle n$:

$\displaystyle \frac{d}{dx}(Ax^{n})=nAx^{n-1}$ 

Applying this rule to each term in the function, we start by taking the first derivative:

$\displaystyle f(x)=6x^{2}+14x-2$

$\displaystyle f'(x)=2(6x^{2-1})+1(14x^{1-1})-0(2)$

$\displaystyle f'(x)=12x+14$

Then, taking the second derivative of the function:

$\displaystyle f'(x)=12x+14$

$\displaystyle f''(x)=1(12x^{1-1})+0(14)$

$\displaystyle f''(x)=12$

We first apply Power Rule.

First Derivative $\displaystyle x^1$:

$\displaystyle 1-1=0$

So result is

$\displaystyle 1x^0$

Anything to a power of $\displaystyle 0$ is $\displaystyle 1$

First Derivative is

$\displaystyle 1$

Second Derivative $\displaystyle 1$:

Any derivative of a constant is $\displaystyle 0$

Second Derivative of $\displaystyle x$ with respect to $\displaystyle x$ is

$\displaystyle 0$

Use Power Rule to take two derivatives of $\displaystyle 7x^5$:

First Derivative:

$\displaystyle 7\cdot5=35$

$\displaystyle 5-1=4$

So result is:

$\displaystyle 35x^4$

Now we take another derivative:

Second Derivative:

$\displaystyle 35\cdot4=140$

$\displaystyle 4-1=3$

So our result is:

$\displaystyle 140x^3$

Apply Power Rule twice.

First Derivative of $\displaystyle x^2$:

$\displaystyle 2\cdot1=2$

$\displaystyle 2-1=1$

So our result is

$\displaystyle 2x^1$

Second Derivative of $\displaystyle x^2$:

$\displaystyle 2\cdot1=2$

$\displaystyle 1-1=0$

So our result is

$\displaystyle 2x^0$

So the second derivative of $\displaystyle x^2$ is

$\displaystyle 2$