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Consider the given function and point.f(x) = -7x4 11x2 - 4, (1, 0)(a) Find an equation of the tangent line to the graph of the function at the given point.

Question

Hi, I am having a hard time figuring out what the equation of the tangent line is in this problem. When I did it I got y=-8x+8 but it's wrong. If someone could help me figure how to get the right answer that would be very helpful!

 

Consider the given function and point.
f(x) = -7x4 11x2 - 4, (1, 0)
(a) Find an equation of the tangent line to the graph of the function at the given point.
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Consider the given function and point. f(x) = -7x4 11x2 - 4, (1, 0) (a) Find an equation of the tangent line to the graph of the function at the given point.

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Step 1

Given:

The functionfis given as f(x)-7x 12-4andthe point is (1,0)
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The functionfis given as f(x)-7x 12-4andthe point is (1,0)

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Step 2

Calculation:

Differentiate the functionfwith respect tox
f(x)7(4x)11(2x)-0
=-28x322x
f'(x) as follows
Substitute thevalue x=1 and obtain the value m =
m f(1
= -28(1)'+ 22(1)
=-28+22
= -6
The slope-intercept form is y = mx +c
help_outline

Image Transcriptionclose

Differentiate the functionfwith respect tox f(x)7(4x)11(2x)-0 =-28x322x f'(x) as follows Substitute thevalue x=1 and obtain the value m = m f(1 = -28(1)'+ 22(1) =-28+22 = -6 The slope-intercept form is y = mx +c

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Step 3

Substitute the ...

For x 1,y 0 and m = -6 in y = mx +c, the value of c becomes
(0) - (-6) (1)+с
0 -6c
—с %3D —6
c 6
Thus, with the help of m and c, the equation of tangent line is obtained
y (6)x+(6)
=-6x +6
help_outline

Image Transcriptionclose

For x 1,y 0 and m = -6 in y = mx +c, the value of c becomes (0) - (-6) (1)+с 0 -6c —с %3D —6 c 6 Thus, with the help of m and c, the equation of tangent line is obtained y (6)x+(6) =-6x +6

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Math

Calculus

Derivative

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