Mixing
Calculus : Plug in value into a derivative?
$begingroup$
$$ f(x) = x^3-2x^2-6x $$
after derivative
$$ 3x^2-4x-6 $$
If u plugin x=1,2,3,4,5
as a result:
$$ 3(1)^2-4(1)-6=-7 $$
$$ 3(2)^2-4(2)-6=-2 $$
$$ 3(3)^2-4(3)-6=9 $$
I know it means the slope of the current point on the curve or instantaneous rate of change
Look at the answer 9 what is it actually mean?Is there any relationshiop between x=3 and x=4? an increment?
calculus
asked Jan 26 at 11:32
Linear Algebra fansLinear Algebra fans
133
$endgroup$
add a comment |
$begingroup$
$$ f(x) = x^3-2x^2-6x $$
after derivative
$$ 3x^2-4x-6 $$
If u plugin x=1,2,3,4,5
as a result:
$$ 3(1)^2-4(1)-6=-7 $$
$$ 3(2)^2-4(2)-6=-2 $$
$$ 3(3)^2-4(3)-6=9 $$
I know it means the slope of the current point on the curve or instantaneous rate of change
Look at the answer 9 what is it actually mean?Is there any relationshiop between x=3 and x=4? an increment?
calculus
asked Jan 26 at 11:32
Linear Algebra fansLinear Algebra fans
133
$endgroup$
1
$begingroup$
The value $25$ is incorrect.
$endgroup$
– Henno Brandsma
Jan 26 at 11:36
4
$begingroup$
Note that $f'(x)=3x^2-4x-6$ and so $f'(3)=9$. This means that at $x=3$, the gradient of the tangent to $f(x)$ is $9$.
$endgroup$
– TheSimpliFire
Jan 26 at 11:36
$begingroup$
$x=3 ; (3)^3-2(3)^2-6(3)=1$ how this answer related to the slope 9?
$endgroup$
– Linear Algebra fans
Jan 26 at 14:20
add a comment |
$begingroup$
$$ f(x) = x^3-2x^2-6x $$
after derivative
$$ 3x^2-4x-6 $$
If u plugin x=1,2,3,4,5
as a result:
$$ 3(1)^2-4(1)-6=-7 $$
$$ 3(2)^2-4(2)-6=-2 $$
$$ 3(3)^2-4(3)-6=9 $$
I know it means the slope of the current point on the curve or instantaneous rate of change
Look at the answer 9 what is it actually mean?Is there any relationshiop between x=3 and x=4? an increment?
calculus
asked Jan 26 at 11:32
Linear Algebra fansLinear Algebra fans
133
$endgroup$
$$ f(x) = x^3-2x^2-6x $$
after derivative
$$ 3x^2-4x-6 $$
If u plugin x=1,2,3,4,5
as a result:
$$ 3(1)^2-4(1)-6=-7 $$
$$ 3(2)^2-4(2)-6=-2 $$
$$ 3(3)^2-4(3)-6=9 $$
I know it means the slope of the current point on the curve or instantaneous rate of change
Look at the answer 9 what is it actually mean?Is there any relationshiop between x=3 and x=4? an increment?
calculus
calculus
asked Jan 26 at 11:32
Linear Algebra fansLinear Algebra fans
133
asked Jan 26 at 11:32
Linear Algebra fansLinear Algebra fans
133
edited Jan 26 at 14:14
Linear Algebra fans
asked Jan 26 at 11:32
Linear Algebra fansLinear Algebra fans
133
asked Jan 26 at 11:32
Linear Algebra fansLinear Algebra fans
133
asked Jan 26 at 11:32
Linear Algebra fansLinear Algebra fans
133
133
1
$begingroup$
The value $25$ is incorrect.
$endgroup$
– Henno Brandsma
Jan 26 at 11:36
4
$begingroup$
Note that $f'(x)=3x^2-4x-6$ and so $f'(3)=9$. This means that at $x=3$, the gradient of the tangent to $f(x)$ is $9$.
$endgroup$
– TheSimpliFire
Jan 26 at 11:36
$begingroup$
$x=3 ; (3)^3-2(3)^2-6(3)=1$ how this answer related to the slope 9?
$endgroup$
– Linear Algebra fans
Jan 26 at 14:20
add a comment |
1
$begingroup$
The value $25$ is incorrect.
$endgroup$
– Henno Brandsma
Jan 26 at 11:36
4
$begingroup$
Note that $f'(x)=3x^2-4x-6$ and so $f'(3)=9$. This means that at $x=3$, the gradient of the tangent to $f(x)$ is $9$.
$endgroup$
– TheSimpliFire
Jan 26 at 11:36
$begingroup$
$x=3 ; (3)^3-2(3)^2-6(3)=1$ how this answer related to the slope 9?
$endgroup$
– Linear Algebra fans
Jan 26 at 14:20
1
1
$begingroup$
The value $25$ is incorrect.
$endgroup$
– Henno Brandsma
Jan 26 at 11:36
$begingroup$
The value $25$ is incorrect.
$endgroup$
– Henno Brandsma
Jan 26 at 11:36
4
4
$begingroup$
Note that $f'(x)=3x^2-4x-6$ and so $f'(3)=9$. This means that at $x=3$, the gradient of the tangent to $f(x)$ is $9$.
$endgroup$
– TheSimpliFire
Jan 26 at 11:36
$begingroup$
Note that $f'(x)=3x^2-4x-6$ and so $f'(3)=9$. This means that at $x=3$, the gradient of the tangent to $f(x)$ is $9$.
$endgroup$
– TheSimpliFire
Jan 26 at 11:36
$begingroup$
$x=3 ; (3)^3-2(3)^2-6(3)=1$ how this answer related to the slope 9?
$endgroup$
– Linear Algebra fans
Jan 26 at 14:20
$begingroup$
$x=3 ; (3)^3-2(3)^2-6(3)=1$ how this answer related to the slope 9?
$endgroup$
– Linear Algebra fans
Jan 26 at 14:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $x=1$ :
$$f(1)=(1)^3-2(1)^2-6(1)=-7$$
$$f'(1)=3(1)^2-4(1)-6=-7$$
This means that at point $(1:,:-7)$ the slope of the tangent is $=-7$ .
For $x=2$ :
$$f(2)=(2)^3-2(2)^2-6(2)=-12$$
$$f'(2)=3(2)^2-4(2)-6=-2$$
This means that at point $(2:,:-12)$ the slope of the tangent is $=-2$ .
For $x=3$ :
$$f(3)=(3)^3-2(3)^2-6(3)=-9$$
$$f'(3)=3(3)^2-4(3)-6=9$$
This means that at point $(3:,:-9)$ the slope of the tangent is $=9$ .
For $x=4$ :
$$f(4)=(4)^3-2(4)^2-6(4)=8$$
$$f'(4)=3(4)^2-4(4)-6=26$$
This means that at point $(4:,:8)$ the slope of the tangent is $=26$ .
answered Jan 26 at 16:58
JJacquelinJJacquelin
45.1k21855
$endgroup$
$begingroup$
Finally I get the answer by myself. The slope mean if x change in a very small amount and y value will change in multiplier of the slope $x^2$ if u put 2 and 2.000001 into x. The slope of 2 is $2(2)=4$ and $2.000001^2=2+0.000001(slope);slope=4$
$endgroup$
– Linear Algebra fans
Jan 27 at 7:25
add a comment |
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$begingroup$
For $x=1$ :
$$f(1)=(1)^3-2(1)^2-6(1)=-7$$
$$f'(1)=3(1)^2-4(1)-6=-7$$
This means that at point $(1:,:-7)$ the slope of the tangent is $=-7$ .
For $x=2$ :
$$f(2)=(2)^3-2(2)^2-6(2)=-12$$
$$f'(2)=3(2)^2-4(2)-6=-2$$
This means that at point $(2:,:-12)$ the slope of the tangent is $=-2$ .
For $x=3$ :
$$f(3)=(3)^3-2(3)^2-6(3)=-9$$
$$f'(3)=3(3)^2-4(3)-6=9$$
This means that at point $(3:,:-9)$ the slope of the tangent is $=9$ .
For $x=4$ :
$$f(4)=(4)^3-2(4)^2-6(4)=8$$
$$f'(4)=3(4)^2-4(4)-6=26$$
This means that at point $(4:,:8)$ the slope of the tangent is $=26$ .
answered Jan 26 at 16:58
JJacquelinJJacquelin
45.1k21855
$endgroup$
$begingroup$
Finally I get the answer by myself. The slope mean if x change in a very small amount and y value will change in multiplier of the slope $x^2$ if u put 2 and 2.000001 into x. The slope of 2 is $2(2)=4$ and $2.000001^2=2+0.000001(slope);slope=4$
$endgroup$
– Linear Algebra fans
Jan 27 at 7:25
add a comment |
$begingroup$
For $x=1$ :
$$f(1)=(1)^3-2(1)^2-6(1)=-7$$
$$f'(1)=3(1)^2-4(1)-6=-7$$
This means that at point $(1:,:-7)$ the slope of the tangent is $=-7$ .
For $x=2$ :
$$f(2)=(2)^3-2(2)^2-6(2)=-12$$
$$f'(2)=3(2)^2-4(2)-6=-2$$
This means that at point $(2:,:-12)$ the slope of the tangent is $=-2$ .
For $x=3$ :
$$f(3)=(3)^3-2(3)^2-6(3)=-9$$
$$f'(3)=3(3)^2-4(3)-6=9$$
This means that at point $(3:,:-9)$ the slope of the tangent is $=9$ .
For $x=4$ :
$$f(4)=(4)^3-2(4)^2-6(4)=8$$
$$f'(4)=3(4)^2-4(4)-6=26$$
This means that at point $(4:,:8)$ the slope of the tangent is $=26$ .
answered Jan 26 at 16:58
JJacquelinJJacquelin
45.1k21855
$endgroup$
$begingroup$
Finally I get the answer by myself. The slope mean if x change in a very small amount and y value will change in multiplier of the slope $x^2$ if u put 2 and 2.000001 into x. The slope of 2 is $2(2)=4$ and $2.000001^2=2+0.000001(slope);slope=4$
$endgroup$
– Linear Algebra fans
Jan 27 at 7:25
add a comment |
$begingroup$
For $x=1$ :
$$f(1)=(1)^3-2(1)^2-6(1)=-7$$
$$f'(1)=3(1)^2-4(1)-6=-7$$
This means that at point $(1:,:-7)$ the slope of the tangent is $=-7$ .
For $x=2$ :
$$f(2)=(2)^3-2(2)^2-6(2)=-12$$
$$f'(2)=3(2)^2-4(2)-6=-2$$
This means that at point $(2:,:-12)$ the slope of the tangent is $=-2$ .
For $x=3$ :
$$f(3)=(3)^3-2(3)^2-6(3)=-9$$
$$f'(3)=3(3)^2-4(3)-6=9$$
This means that at point $(3:,:-9)$ the slope of the tangent is $=9$ .
For $x=4$ :
$$f(4)=(4)^3-2(4)^2-6(4)=8$$
$$f'(4)=3(4)^2-4(4)-6=26$$
This means that at point $(4:,:8)$ the slope of the tangent is $=26$ .
answered Jan 26 at 16:58
JJacquelinJJacquelin
45.1k21855
$endgroup$
For $x=1$ :
$$f(1)=(1)^3-2(1)^2-6(1)=-7$$
$$f'(1)=3(1)^2-4(1)-6=-7$$
This means that at point $(1:,:-7)$ the slope of the tangent is $=-7$ .
For $x=2$ :
$$f(2)=(2)^3-2(2)^2-6(2)=-12$$
$$f'(2)=3(2)^2-4(2)-6=-2$$
This means that at point $(2:,:-12)$ the slope of the tangent is $=-2$ .
For $x=3$ :
$$f(3)=(3)^3-2(3)^2-6(3)=-9$$
$$f'(3)=3(3)^2-4(3)-6=9$$
This means that at point $(3:,:-9)$ the slope of the tangent is $=9$ .
For $x=4$ :
$$f(4)=(4)^3-2(4)^2-6(4)=8$$
$$f'(4)=3(4)^2-4(4)-6=26$$
This means that at point $(4:,:8)$ the slope of the tangent is $=26$ .
answered Jan 26 at 16:58
JJacquelinJJacquelin
45.1k21855
answered Jan 26 at 16:58
JJacquelinJJacquelin
45.1k21855
answered Jan 26 at 16:58
JJacquelinJJacquelin
45.1k21855
answered Jan 26 at 16:58
JJacquelinJJacquelin
45.1k21855
45.1k21855
$begingroup$
Finally I get the answer by myself. The slope mean if x change in a very small amount and y value will change in multiplier of the slope $x^2$ if u put 2 and 2.000001 into x. The slope of 2 is $2(2)=4$ and $2.000001^2=2+0.000001(slope);slope=4$
$endgroup$
– Linear Algebra fans
Jan 27 at 7:25
add a comment |
$begingroup$
Finally I get the answer by myself. The slope mean if x change in a very small amount and y value will change in multiplier of the slope $x^2$ if u put 2 and 2.000001 into x. The slope of 2 is $2(2)=4$ and $2.000001^2=2+0.000001(slope);slope=4$
$endgroup$
– Linear Algebra fans
Jan 27 at 7:25
$begingroup$
Finally I get the answer by myself. The slope mean if x change in a very small amount and y value will change in multiplier of the slope $x^2$ if u put 2 and 2.000001 into x. The slope of 2 is $2(2)=4$ and $2.000001^2=2+0.000001(slope);slope=4$
$endgroup$
– Linear Algebra fans
Jan 27 at 7:25
$begingroup$
Finally I get the answer by myself. The slope mean if x change in a very small amount and y value will change in multiplier of the slope $x^2$ if u put 2 and 2.000001 into x. The slope of 2 is $2(2)=4$ and $2.000001^2=2+0.000001(slope);slope=4$
$endgroup$
– Linear Algebra fans
Jan 27 at 7:25
add a comment |
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1
$begingroup$
The value $25$ is incorrect.
$endgroup$
– Henno Brandsma
Jan 26 at 11:36
4
$begingroup$
Note that $f'(x)=3x^2-4x-6$ and so $f'(3)=9$. This means that at $x=3$, the gradient of the tangent to $f(x)$ is $9$.
$endgroup$
– TheSimpliFire
Jan 26 at 11:36
$begingroup$
$x=3 ; (3)^3-2(3)^2-6(3)=1$ how this answer related to the slope 9?
$endgroup$
– Linear Algebra fans
Jan 26 at 14:20