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At what points on the graph of #f(x)=2x^3-9x^2-12x+5# is the slope of the tangent line 12?

Answer 1

Samantha Adkison

The derivative of a function gives us the slope of a tangent line for a specified value of #x#

#f(x) = 2x^3-9x^2-12x+5# implies #f'(x) = 6x^2-18x-12#

and we are asked to find when this is #=12#

#6x^2-18x-12 = 12#

#x^2-3x -4=0#

which factors as #(x-4)(x+1)=0#

when #x=4# #f(x) = 2(4)^3-9(4)^2-12(4)+5# #=-59#

when #x=-1# #f(-1)=2(-1)^3-9(-1)^2-12(-1)+5# #=6#

So #f(x)# has a tangent with slope #12# at #(4,-59)# and #(-1,6)#

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Answer 2

Axel Alvarez

To find the points on the graph where the slope of the tangent line is 12, we need to find the derivative of the function f(x) and then set it equal to 12.

The derivative of f(x) is f'(x) = 6x^2 - 18x - 12.

Setting f'(x) equal to 12, we have:

6x^2 - 18x - 12 = 12.

Now, rearrange and solve for x:

6x^2 - 18x - 24 = 0.

Now, solve this quadratic equation to find the values of x. Once you have the values of x, plug them back into the original function f(x) to find the corresponding y-values. These (x, y) points represent where the slope of the tangent line is 12 on the graph of f(x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.