Show that the curve y = 2*e^x +3x + 5x^3 has no tangent line with slope 2.

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We have to show that the curve `y=2e^x+3x+5x^3`  has no tangent line with slope 2.

Now, we know that the slope of the tangent is found by taking the first derivative of the curve. In other words,

`\frac{dy}{dx}=2e^x+3+15x^2`

Now, let us consider that there is a tangent line whose slope...

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We have to show that the curve `y=2e^x+3x+5x^3`  has no tangent line with slope 2.

Now, we know that the slope of the tangent is found by taking the first derivative of the curve. In other words,

`\frac{dy}{dx}=2e^x+3+15x^2`

Now, let us consider that there is a tangent line whose slope is 2.

To demonstrate,

`\frac{dy}{dx}=2`

`2e^x+3+15x^2=2`

For example, `2e^x+15x^2=-1`

But we know that `e^x ` and `x^2` are always positive. So the left hand side of the above equation will be always positive, but the right hand side is always negative.

Hence, there are no values of x for which we get the slope 2.

Therefore we can say that there are no tangent lines to the given curve with a slope of 2.

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We are asked to show that curve represented by the function `y=2e^x+3x+5x^3 ` has no tangent line with a slope of 2:

The slope of a tangent line to a curve at a point is found by evaluating the first derivative of the function at that point.

Here the first derivative is:

`y'=2e^x+3+15x^2 `

If we temporarily assume that there is a tangent line with slope 2, we can try to find the x-coordinate(s) of the point(s) where the value of the derivative is 2.

`y'=2 ==> 2e^x+3+15x^2=2 `

Then `2e^x=-1-15x^2 `

Note that the left-hand side of the equation is always positive, while the right-hand side is always negative. Thus there are no real values of x such that the value of the derivative is 2.

The graph:

 

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As discussed above, the slope of the line tangent to a curve at a given point equals the value of the derivative of the function describing the curve at this point. So, for 

`y = 2e^x + 3x + 5x^3` the slope of the tangent line at the point `x_0` will be:

`m = y'(x_0) = 2e^(x_0) + 3 + 15x_0^2` .

Let's analyze this expression. The first term is the exponential function of x, which is always positive (the domain of `e^x` is all positive numbers). The last term is 15 times  `x_0^2` , which is nonnegative (can be positive or zero). Consequently, the sum of the first and third terms is a number always greater than 0.

If the second term, 3, is added to the sum of the first and third terms, the result has to be always greater than 3.

This means has to be always greater than 3, so it cannot possibly be equal to 2.

Therefore, the curve with the given equation has no tangent line with slope 2.

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The slope of a line tangent to the curve y = f(x) at a point where x = a, is given by the value of f'(a).

In the problem, the curve is represented by y = 2*e^x + 5x^3 + 3x. First determine the derivative `(dy)/(dx)` .

`(dy)/(dx)= (2*e^x + 5x^3 + 3x)'`

= `2*e^x + 15x^2 + 3`

If a tangent to this curve has slope 2, the equation 2*e^x + 15x^2 + 3 = 2 should have a real root.

2*e^x + 15x^2 + 3 = 2

=> 2*e^x + 15x^2 = 2 - 3 = -1

As e is positive, for no real value of x is the value of e^x negative. Similarly, x^2 is positive for all real values of x. The value of 2*e^x + 15x^2 is positive for all real values of x.

Therefore, the equation 2*e^x + 15x^2 = -1 does not have a real root. As a result, the slope of a line tangent to the given curve cannot be equal to 2.

This proves that the curve y = 2*e^x +3x + 5x^3 has no tangent line with slope 2.

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