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Curve Sketching 1Curve Sketching 1
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Question 1 of 5
1. Question
Sketch the First Derivative Curve of the curve shown below:
Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA positive `f'(x)` means a positive slopeA negative `f'(x)` means a negative slopeFirst, set up a separate plane with the `y` axis replaced with `f'(x)` axisNext, mark out the stationary points from the given curve into the revised graphNote that in the revised graph, stationary points are always on the `x` axis
For an easier transfer, mark out the movement of the given curve
Remember that a positive or increasing slope means `f'(x)` is positive, and a negative or decreasing slope means `f'(x)` is negativeFinally, draw a curve on the revised graph depending on the signs marked on the original curve
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Question 2 of 5
2. Question
Sketch the curve:`f(x)=x^2-2x`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA positive `f'(x)` means a positive slopeA negative `f'(x)` means a negative slopeFirst, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `x^2-2x` `f'(x)` `=` `2x-2` `2x-2` `=` `0` Equate to `0` `2(x-1)` `=` `0` Factor out `2` `x` `=` `1` This means that there is a stationary point at `x=1`Find the corresponding `y` value of `x=1``f(x)` `=` `x^2-2x` $$f(\color{#007DDC}{1})$$ `=` $$\color{#007DDC}{1}^2)-2(\color{#007DDC}{1})$$ `=` `1-2` `=` `-1` Stationary Point: $$(\color{#007DDC}{1},\color{#e65021}{-1})$$
Next, graph `f'(x)=2x-2` to identify where the curve increases or decreasesStart by setting up a plane with the `y` axis replaced with `f'(x)` axis and plot `x=1`
Substitute `x=0` to know where the derivative graph intersects the `f'(x)` axis`f'(x)` `=` `2x-2` `=` `2(0)-2` Substitute `x=0` `=` `0-2` `f'(x)` `=` `-2` 
Form a line by connecting the two points
From the diagram above, we can say the following about the curve of `f(x)`:The slope is negative or decreasing at `x≤1`The slope is positive or increasing at `x≥1`Next, check where the curve intersects the `x` axis by equating `f(x)` to `0``f(x)` `=` `0` `x^2-2x` `=` `0` Equate to `0` `x(x-2)` `=` `0` `x` `=` `0,2` The curve intersects `x=0` and `x=2`
Lastly, draw a curve or a parabola over the plotted points, following the increase and decrease that have been set with the derivative graph
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Question 3 of 5
3. Question
Sketch the curve`f(x)=x^4-8x^2+16`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA sign diagram is a way of visualizing a curve’s graph by indicating the inflection points, stationary points, and the increases or decreases in the curve.First, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `x^4-8x^2+16` `f'(x)` `=` `4x^3-16x=0` Equate `f'(x)` to `0` `=` `4x(x^2-4)=0` Factor out `4x` `=` `4x(x-2)(x+2)=0` Factor out the difference of `2` squares `x=0,x=2,x=-2` Find the corresponding `y` values by substituting each `x` value to the function`f(x)` `=` `x^4-8x^2+16` $$f(\color{#007DDC}{0})$$ `=` $$\color{#007DDC}{0}^4-8(\color{#007DDC}{0}^2)+16$$ `=` `0-0+16` `=` `16` Stationary Point: $$(\color{#007DDC}{0},\color{#e65021}{16})$$`f(x)` `=` `x^4-8x^2+16` $$f(\color{#007DDC}{2})$$ `=` $$\color{#007DDC}{2}^4-8(\color{#007DDC}{2}^2)+16$$ `=` `16-32+16` `=` `0` Stationary Point:
$$(\color{#007DDC}{2},\color{#e65021}{0})$$`f(x)` `=` `x^4-8x^2+16` $$f(\color{#007DDC}{-2})$$ `=` $$\color{#007DDC}{-2}^4-8(\color{#007DDC}{-2}^2)+16$$ `=` `16-32+16` `=` `0` Stationary Point:
$$(\color{#007DDC}{-2},\color{#e65021}{0})$$
Next, create a sign diagram to identify where the curve increases or decreasesStart by setting up a horizontal line with matching indicators of the stationary points
Test the gradient of a point to the left of `-2` such as `x=-3``f'(x)` `=` `4x^3-16x` $$f'(\color{#00880A}{-3})$$ `=` $$4(\color{#00880A}{-3})^3-16(\color{#00880A}{-3})$$ `=` `4(-27)+48` `=` `-108+48` `=` `-60` This value is negative, which means the curve’s slope at `-3` is decreasingIndicate this on the sign diagram by adding a negative sign to the left of `-2`
Test the gradient of a point to the right of `-2` such as `x=-1``f'(x)` `=` `4x^3-16x` $$f'(\color{#00880A}{-1})$$ `=` $$4(\color{#00880A}{-1})^3-16(\color{#00880A}{-1})$$ `=` `4(-1)+16` `=` `-4+16` `=` `12` This value is positive, which means the curve’s slope at `-1` is increasingIndicate this on the sign diagram by adding a positive sign to the right of `-2`
Test the gradient of a point to the left of `2` such as `x=1``f'(x)` `=` `4x^3-16x` $$f'(\color{#00880A}{1})$$ `=` $$4(\color{#00880A}{1}^3)-16(\color{#00880A}{1})$$ `=` `4(1)-16` `=` `4-16` `=` `-12` This value is negative, which means the curve’s slope at `1` is decreasingIndicate this on the sign diagram by adding a negative sign to the left of `2`
Test the gradient of a point to the right of `2` such as `x=3``f'(x)` `=` `4x^3-16x` $$f'(\color{#00880A}{3})$$ `=` $$4(\color{#00880A}{3}^3)-16(\color{#00880A}{3})$$ `=` `4(27)-48` `=` `108-48` `=` `60` This value is positive, which means the curve’s slope at `3` is increasingIndicate this on the sign diagram by adding a positive sign to the left of `2`
Finally, draw a curve along the stationary points with the help of the sign diagram
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Question 4 of 5
4. Question
To help identify where the curve increases or decreases, sketch the first derivative of the curve:`f(x)=x^3-6x^2+9x+1`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA positive `f'(x)` means a positive slopeA negative `f'(x)` means a negative slopeFirst, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `x^3-6x^2+9x+1` `f'(x)` `=` `3x^2-12x+9` `3x^2-12x+9` `=` `0` Equate to `0` `3(x^2-4x+3)` `=` `0` Factor out `3` `3(x-3)(x-1)` `=` `0` Factor out further `x` `=` `3,1` This means that there are stationary points at `x=3` and `x=1`To start graphing the derivative, set up a plane with the `y` axis replaced with `f'(x)` axis
Plot `x=1` and `x=3` since this is where the derivative would intersect the `x` axis
Recall that `f'(x)=3x^2-12x+9`Since the highest power is `2`, the graph would be a parabola. Draw this over the two plotted points
From the diagram above, we can say the following about the curve of `f(x)`:The slope is positive at `x≤1`The slope is negative at `1≤x≤3`The slope is positive at `x≥3`
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Question 5 of 5
5. Question
Sketch the curve:`f(x)=x^3-6x^2+9x+1`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA positive `f'(x)` means a positive slopeA negative `f'(x)` means a negative slopeFirst, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `x^3-6x^2+9x+1` `f'(x)` `=` `3x^2-12x+9` `3x^2-12x+9` `=` `0` Equate to `0` `3(x^2-4x+3)` `=` `0` Factor out `3` `3(x-3)(x-1)` `=` `0` Factor out further `x` `=` `3,1` This means that there are stationary points at `x=3` and `x=1`Find the corresponding `y` values by substituting each `x` value to the function`f(x)` `=` `x^3-6x^2+9x+1` $$f(\color{#007DDC}{3})$$ `=` $$\color{#007DDC}{3}^3-6(\color{#007DDC}{3}^2)+9(\color{#007DDC}{3})+1$$ `=` `27-54+27+1` `=` `1` Stationary Point: $$(\color{#007DDC}{3},\color{#e65021}{1})$$`f(x)` `=` `x^3-6x^2+9x+1` $$f(\color{#007DDC}{1})$$ `=` $$\color{#007DDC}{1}^3-6(\color{#007DDC}{1}^2)+9(\color{#007DDC}{1})+1$$ `=` `1-6+9+1` `=` `5` Stationary Point: $$(\color{#007DDC}{1},\color{#e65021}{5})$$
Next, graph `f'(x)=3x^2-12x+9` to identify where the curve increases or decreasesStart by setting up a plane with the `y` axis replaced with `f'(x)` axis and plot `x=1` and `x=3`
Since the highest power of `f'(x)` is `2`, draw a parabola over the plotted points
From the diagram above, we can say the following about the curve of `f(x)`:The slope is positive or increasing at `x≤1`The slope is negative or decreasing at `1≤x≤3`The slope is positive or increasing at `x≥3`Next, check where the curve intersects the `y` axis by substituting `x=0` to `f(x)``f(x)` `=` `x^3-6x^2+9x+1` `f(0)` `=` `0^3-6(0^2)+9(0)+1` Substitute `x=0` `=` `0-0+0+1` `y` `=` `1` The curve intersects `y=1`
Lastly, draw a curve over the plotted points, following the increase and decrease that have been set with the derivative graph