Set the derivative equal to ( 0) then solve the equation( 3x^2-14x=0). ( x=0,(14)/3) Solve the original function( f(x)=x^3-7x^2) at ( x=0). ( 0) Solve the original function( f(x)=x^3-7x^2) at ( x=(14)/3). ( -(1372)/(27)) The horizontaltangentl...
Subtract ( xy) from both sides of the equation. ( (sin)(x+y)-xy=0) Set( (sin)(x+y)-xy) as a function of ( x). ( f(x)=0) ( f=0) is not a function of ( f=0). Therefore, it is either a horizontal or verticalfunction and has no horizontaltangentline. No horizo...
( f(x)=2x) Find the derivative. ( 2) Since ( 2≠ 0), there are no solutions. No solution There are no solution found by setting the derivative equal to ( 0), ( 2=0) so there are no horizontaltangentlines. No horizontaltangentlines found反馈...
Note that both derivatives (dy/dθ and dx/dθ) are 0 when θ=0. Using this information alone, you do not know whether the graph has a horizontal or vertical tangent line at the pole. From Figure, however, you can see that the graph has a cusp at the pole....
Find the points on the given curve where the tangent line is horizontal or vertical. r=e^(θ ) 相关知识点: 试题来源: 解析 horizontal tangents at (e^(π (n- 1/4)),π (n- 14)).vertical tangents at (e^(π (n+ 1/4)),π (n+ 14)). r=e^(θ ) ⇒ x=rcos θ=e^(θ...
Find the points on the given curve where the tangent line is horizontal or vertical. r=e^(θ ) 相关知识点: 试题来源: 解析 horizontal tangents at (e^(π (n- 1/4)),π (n- 14)). vertical tangents at (e^(π (n+ 1/4)),π (n+ 14))....
(ii) above are satisfied, we should compute the derivative. y = 1 2 x −1/2 − 3 2 x 1/2 = 1 2 √ x − 3 √ x 2 = 1 2 √ x − 3x 2 √ x = 1 −3x 2 √ x . Therefore, when x = 1 3 , y = 0 and so y = f(x) has a horizontal tangent line ...
Find the value on the horizontal axis or x value of the point of the curve you want to calculate the tangent for and replace x on the derivative function by that value. To calculate the tangent of the example function at the point where x = 2, the resulting value would be f'(2) =...
Solve for the function with the value for x you just inserted. The example function is 12(9) + 2 = 110. This is the slope of the tangent line to the original function at that x value. TL;DR (Too Long; Didn't Read) Because the tangent line will be horizontal at a maximum or mi...
Find the points on the curve where the tangent line is horizontal. (a) x = 5 ( cos ? ? cos 2 ? ) , y = 5 ( sin ? ? sin ? cos ? ) Find the slope of the tangent line to the given polar curve at the Find equation of the tangent line...