Determine the critical points for each function. Determine...

Determine the critical points for each function. Determine whether the critical point is a local maximum or minimum, and whether or not the tangent is parallel to the horizontal axis. a. $h(x)=-6 x^{3}+18 x^{2}+3 \quad$ c. $y=(x-5)^{\frac{1}{5}}$ b. $g(t)=t^{5}+t^{3}$ d. $f(x)=\left(x^{2}-1\right)^{\frac{1}{3}}$

Transcript

00:01 Hi there.

00:01 So for this question we have to determine the critical point so for each function and let's determine this critical point is a local maximum minimum and or an and the tangent is parallel to the horizontal x -axis that's it.

00:19 So first of all let's start with part a in the part a we have the function let's say which one age of x yes age of x is equal to negative 6 x cubed and plus 18 x squared and plus 3.

00:35 So for the critical point, we have to just take the derivative of the function, which is negative 18x squared and plus 36x and set this equation equal to 0.

00:47 So 18x is the common term negative x plus 2, which is equal to 0.

00:52 So if i just set each factor equal to 0, so i can get 18 x is equal to 0, that means x is equal to 0.

01:00 Negative x plus 2 is equal to 0 that means x is equal to 2 so we have two points i'm going to just write orderly and just make a sign table here for the derivative of the function which is edge derivative so if i just take a look at the leading term for the derivative function which is negative 18 that means i'm going to start with negative on the right hand side of the table negative plus and negative by the way x is equal to 0 and x equal to two are both single terms single roots so that means the function is decreasing here increasing and decreasing that means for x is equal to zero we have a minimum point minimum point and for x is equal to two we got a maximum point so we we both have minimum and maximum point in this question so the tangent is parallel to the horizontal x -axis.

02:11 So if i just draw a tangent at x -s -equal to 0, which is parallel to x -axis, why? because if you plug in the derivative of the function with 0, we will get 0.

02:31 That means the slope is 0, so it is parallel to the x -axis.

02:36 And also the tangent at x -6.

02:40 Is equal to 2 is also parallel to x -axis because the derivative of the function at x is equal to 2 is again so this is for part a what about part b so in part b we have different function which is g of d and that is equal to t to the fifth plus t to the third again we are going to follow the same process let's first of all take the derivative of the function which is five t to the fourth plus three t the second and set this equation equal to zero so take t squared parentheses which is 5 t squared and plus one no plus one this is plus three and that is equal to zero so from this case i'm going to set each factor equal to zero so t square is equal to zero that means t is equal to zero but be careful the power of the t is an even number which is two so that means this is a double root.

03:47 And for 5 t squared plus 3 is equal to 0, that means t square is equal to negative 3 over 5...

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